面试刷题

Binary Tree based Recursion & Divide Conquer 二叉树递归与分治

226. Invert Binary Tree (Easy)

Invert a binary tree.

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Example:
Input:
4
/ \
2 7
/ \ / \
1 3 6 9
Output:
4
/ \
7 2
/ \ / \
9 6 3 1

Trivia:
This problem was inspired by this original tweet by Max Howell:
Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so f*** off.

递归1:

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class Solution:
def invertTree(self, root):
if not root:
return
root.left, root.right = root.right, root.left
self.invertTree(root.left)
self.invertTree(root.right)
return root

递归2:

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class Solution:
def invertTree(self, root):
if not root:
return
l = self.invertTree(root.left)
r = self.invertTree(root.right)
root.left, root.right = r, l
return root

非递归:

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class Solution:
def invertTree(self, root):
if not root:
return
stack = [root]
while stack:
cur = stack.pop(-1)
cur.left, cur.right = cur.right, cur.left
if cur.left:
stack.append(cur.left)
if cur.right:
stack.append(cur.right)
return root

九刷:高频

100. Same Tree (Easy)

Given two binary trees, write a function to check if they are the same or not.
Two binary trees are considered the same if they are structurally identical and the nodes have the same value.

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Example 1:
Input: 1 1
/ \ / \
2 3 2 3

[1,2,3], [1,2,3]
Output: true

Example 2:
Input: 1 1
/ \
2 2

[1,2], [1,null,2]
Output: false

Example 3:
Input: 1 1
/ \ / \
2 1 1 2

[1,2,1], [1,1,2]
Output: false

递归:

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class Solution:
def isSameTree(self, p: TreeNode, q: TreeNode) -> bool:
if not p and not q:
return True
if (p and not q) or (q and not p) or p.val != q.val:
return False
return self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right)

非递归:

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class Solution:
def isSameTree(self, p: TreeNode, q: TreeNode) -> bool:
if not p and not q:
return True
stack = [(p, q)]
while stack:
p, q = stack.pop(-1)
if not p and not q:
continue
if (p and not q) or (q and not p) or p.val != q.val:
return False
stack.append((p.left, q.left))
stack.append((p.right, q.right))
return True

五刷:高频

101. Symmetric Tree (Easy)

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

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For example, this binary tree [1,2,2,3,4,4,3] is symmetric:
1
/ \
2 2
/ \ / \
3 4 4 3
But the following [1,2,2,null,3,null,3] is not:
1
/ \
2 2
\ \
3 3

Note:
Bonus points if you could solve it both recursively and iteratively.

递归

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class Solution:
def isSymmetric(self, root: TreeNode) -> bool:
if not root:
return True
def helper(l, r):
if not l and not r:
return True
if (l and not r) or (not l and r):
return False
return l.val == r.val and helper(l.left, r.right) and helper(l.right, r.left)
return helper(root.left, root.right)

非递归

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class Solution:
def isSymmetric(self, root: TreeNode) -> bool:
if not root:
return True
q = [(root.left, root.right)]
while q:
l, r = q.pop(0)
if (l and not r) or (r and not l):
return False
if (l and r) and (l.val != r.val):
return False
if l:
q.append((l.left, r.right))
q.append((l.right, r.left))
return True

六刷:高频

104. Maximum Depth of Binary Tree (Easy)

again!
Given a binary tree, find its maximum depth.
The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.
Note: A leaf is a node with no children.

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Example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its depth = 3.

递归1:

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class Solution:
def maxDepth(self, root):
if not root:
return 0
ans = 0
def helper(root, cur):
nonlocal ans
if not root:
return
if not root.left and not root.right:
ans = max(ans, cur)
helper(root.left, cur + 1)
helper(root.right, cur + 1)
helper(root, 1)
return ans

非递归:

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class Solution:
def maxDepth(self, root):
if not root:
return 0
ans = 0
stack = [(root, 1)]
while stack:
root, cur = stack.pop()
if cur > ans:
ans = cur
if root.left:
stack.append((root.left, cur + 1))
if root.right:
stack.append((root.right, cur + 1))
return ans

七刷:高频

111. Minimum Depth of Binary Tree (Easy)

again!
Given a binary tree, find its minimum depth.
The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.
Note: A leaf is a node with no children.
Example:

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Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its minimum depth = 2.

递归

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class Solution:
def minDepth(self, root: TreeNode) -> int:
if not root:
return 0
ans = sys.maxsize
def helper(root, cur):
nonlocal ans
if not root:
return
if not root.left and not root.right:
ans = min(ans, cur)
helper(root.left, cur + 1)
helper(root.right, cur + 1)
helper(root, 1)
return ans

非递归

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class Solution:
def minDepth(self, root: TreeNode) -> int:
if not root:
return 0
q = [(root, 1)]
while q:
n, depth = q.pop(0)
if not n.left and not n.right:
return depth
if n.left:
q.append((n.left, depth + 1))
if n.right:
q.append((n.right, depth + 1))

六刷:高频

144. Binary Tree Preorder Traversal (Medium)

again!
Given a binary tree, return the preorder traversal of its nodes’ values.

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Example:
Input: [1,null,2,3]
1
\
2
/
3

Output: [1,2,3]

Follow up: Recursive solution is trivial, could you do it iteratively?

递归 Divide and Conquer 分治 / backtracking:

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class Solution:
def preorderTraversal(self, root):
if not root:
return []
l = self.preorderTraversal(root.left)
r = self.preorderTraversal(root.right)
return [root.val] + l + r

非递归:

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class Solution:
def preorderTraversal(self, root):
if not root:
return []
ans = []
stack = []
while stack or root:
if root:
ans.append(root.val)
if root.right:
stack.append(root.right)
root = root.left
else:
root = stack.pop(-1)
return ans

递归 Traverse:

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class Solution:
def preorderTraversal(self, root):
ans = []
def helper(root):
if not root:
return
ans.append(root.val)
helper(root.left)
helper(root.right)
helper(root)
return ans

七刷

94. Binary Tree Inorder Traversal (Medium)

Given a binary tree, return the inorder traversal of its nodes’ values.
Example:

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Input: [1,null,2,3]
1
\
2
/
3

Output: [1,3,2]

Follow up: Recursive solution is trivial, could you do it iteratively?

递归 Traverse:

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class Solution:
def inorderTraversal(self, root):
ans = []
def helper(root):
if not root:
return
helper(root.left)
ans.append(root.val)
helper(root.right)
helper(root)
return ans

非递归:

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class Solution:
def inorderTraversal(self, root):
if not root:
return []
ans = []
stack = []
while stack or root:
if root:
stack.append(root)
root = root.left
else:
root = stack.pop(-1)
ans.append(root.val)
root = root.right
return ans

递归 Divide and Conquer/backtracking 分治:

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class Solution:
def inorderTraversal(self, root):
if not root:
return []
l = self.inorderTraversal(root.left)
r = self.inorderTraversal(root.right)
return l + [root.val] + r

八刷:高频, 非递归,stack初始为空,while root or stack: if root: 向左入栈到底 else: root出栈,root.val入ans,向右走…root = root.right

230. Kth Smallest Element in a BST (Medium)

Given a binary search tree, write a function kthSmallest to find the kth smallest element in it.
Note:
You may assume k is always valid, 1 ≤ k ≤ BST’s total elements.

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Example 1:
Input: root = [3,1,4,null,2], k = 1
3
/ \
1 4
\
2
Output: 1

Example 2:
Input: root = [5,3,6,2,4,null,null,1], k = 3
5
/ \
3 6
/ \
2 4
/
1
Output: 3

Follow up:
What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently? How would you optimize the kthSmallest routine?

递归:

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class Solution:
def kthSmallest(self, root: TreeNode, k: int) -> int:
ans = None
def helper(root):
nonlocal k, ans
if not root:
return
helper(root.left)
k -= 1
if k == 0:
ans = root.val
return
helper(root.right)
helper(root)
return ans

非递归:

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class Solution:
def kthSmallest(self, root: TreeNode, k: int) -> int:
cnt = 0
stack = []
while stack or root:
if root:
stack.append(root)
root = root.left
else:
root = stack.pop(-1)
cnt += 1
if cnt == k:
return root.val
root = root.right

七刷:面经,Triplebyte, 维萨
Follow up: 二叉树经常被修改 如何优化 kthSmallest 这个操作? leetcode官方答案提出做一个类似LRU双链表的数据结构实现O(h + k)插删和查询

129. Sum Root to Leaf Numbers (Medium)

again!
Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.
An example is the root-to-leaf path 1->2->3 which represents the number 123.
Find the total sum of all root-to-leaf numbers.
Note: A leaf is a node with no children.

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Example:
Input: [1,2,3]
1
/ \
2 3
Output: 25
Explanation:
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.
Therefore, sum = 12 + 13 = 25.

Example 2:
Input: [4,9,0,5,1]
4
/ \
9 0
/ \
5 1
Output: 1026
Explanation:
The root-to-leaf path 4->9->5 represents the number 495.
The root-to-leaf path 4->9->1 represents the number 491.
The root-to-leaf path 4->0 represents the number 40.
Therefore, sum = 495 + 491 + 40 = 1026.

递归:

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class Solution:
def sumNumbers(self, root: TreeNode) -> int:
if not root:
return 0
ans = 0
def helper(root, cur):
nonlocal ans
if not root:
return
cur += root.val
if not root.left and not root.right:
ans += cur
return
helper(root.left, cur * 10)
helper(root.right, cur * 10)
helper(root, 0)
return ans

非递归:

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class Solution:
def sumNumbers(self, root: TreeNode) -> int:
if not root:
return 0
ans = 0
s = [(root, 0)]
while s:
root, cur = s.pop(-1)
if root:
cur += root.val
if not root.left and not root.right:
ans += cur
continue
s.append((root.left, cur * 10))
s.append((root.right, cur * 10))
return ans

七刷:高频

116. Populating Next Right Pointers in Each Node (Medium)

You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:

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struct Node {
int val;
Node *left;
Node *right;
Node *next;
}

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.
Initially, all next pointers are set to NULL.
Example:
Populating Next Right Pointers example

递归:

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class Solution:
def connect(self, root: 'Node') -> 'Node':
if not root:
return root
if root.left:
root.left.next = root.right
if root.next:
root.right.next = root.next.left
self.connect(root.right)
self.connect(root.left)
return root

六刷:高频,非递归允许用额外空间的话比较简单,就不写了,用O(1)额外空间的非递归写法与117题差不多

117. Populating Next Right Pointers in Each Node II (Medium)

again!
Given a binary tree

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struct Node {
int val;
Node *left;
Node *right;
Node *next;
}

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.
Initially, all next pointers are set to NULL.
Example:
Populating Next Right Pointers example II
Note:
You may only use constant extra space.
Recursive approach is fine, implicit stack space does not count as extra space for this problem.

递归:

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class Solution:
def connect(self, root: 'Node') -> 'Node':
if not root:
return
def helper(root):
if not root:
return
if root.left:
return root.left
if root.right:
return root.right
return helper(root.next)
if root.left:
if root.right:
root.left.next = root.right
else:
root.left.next = helper(root.next)
if root.right:
root.right.next = helper(root.next)
self.connect(root.right)
self.connect(root.left)
return root

非递归:

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class Solution:
def connect(self, root: 'Node') -> 'Node':
if not root:
return
cur = root
nextLevelHead = Node(0)
while cur:
p = nextLevelHead
while cur:
if cur.left:
p.next = cur.left
p = p.next
if cur.right:
p.next = cur.right
p = p.next
cur = cur.next
cur = nextLevelHead.next
nextLevelHead.next = None
return root

六刷:高频

110. Balanced Binary Tree (Easy)

again!
Given a binary tree, determine if it is height-balanced.
For this problem, a height-balanced binary tree is defined as:
a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

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Example 1:
Given the following tree [3,9,20,null,null,15,7]:
3
/ \
9 20
/ \
15 7
Return true.

Example 2:
Given the following tree [1,2,2,3,3,null,null,4,4]:
1
/ \
2 2
/ \
3 3
/ \
4 4
Return false.

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class Solution:
def isBalanced(self, root):
if not root:
return True
ans = True
def helper(root):
nonlocal ans
if not root:
return 0
l = helper(root.left)
r = helper(root.right)
if abs(l - r) > 1:
ans = False
return max(l, r) + 1
helper(root)
return ans

九刷:高频。非递归需要用postorder, 比较难,以后写,TODO

108. Convert Sorted Array to Binary Search Tree (Easy)

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:
Given the sorted array: [-10,-3,0,5,9],
One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

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5
     0
/ \
-3 9
/ /
-10 5

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class Solution:
def sortedArrayToBST(self, nums: List[int]) -> TreeNode:
if not nums:
return
rIdx = len(nums) // 2
root = TreeNode(nums[rIdx])
root.left = self.sortedArrayToBST(nums[:rIdx])
root.right = self.sortedArrayToBST(nums[rIdx + 1:])
return root

六刷:高频, 面经, Amazon。无他手熟尔 todo:遍历方法

109. Convert Sorted List to Binary Search Tree (Medium)

again!
Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:
Given the sorted linked list: [-10,-3,0,5,9],
One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

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5
     0
/ \
-3 9
/ /
-10 5

解法1:space: O(n):

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class Solution:
def sortedListToBST(self, head: ListNode) -> TreeNode:
if not head:
return None
arr = []
while head:
arr.append(head.val)
head = head.next
def helper(nums):
if not nums:
return
rIdx = len(nums) // 2
root = TreeNode(nums[rIdx])
root.left = helper(nums[:rIdx])
root.right = helper(nums[rIdx + 1:])
return root
return helper(arr)

解法2:space: O(1):

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class Solution:
def sortedListToBST(self, head: ListNode) -> TreeNode:
if h == t:
return
s, f = h, h
while f != t and f.next != t:
s = s.next
f = f.next.next
root = TreeNode(s.val)
root.left = helper(h, s)
root.right = helper(s.next, t)
return root
return helper(head, None)

五刷:高频,time complexity: O(n), O(nlogn)

112. Path Sum (Easy)

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
Note: A leaf is a node with no children.

Example:
Given the below binary tree and sum = 22,

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      5
/ \
4 8
/ / \
11 13 4
/ \ \
7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

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class Solution:
def hasPathSum(self, root: TreeNode, sum: int) -> bool:
if not root:
return False
ans = False
def helper(root, cur):
nonlocal ans
if not root:
return
cur -= root.val
if not root.left and not root.right:
if cur == 0:
ans = True
return
helper(root.left, cur)
helper(root.right, cur)
helper(root, sum)
return ans

七刷:高频,面经,Quora

199. Binary Tree Right Side View (Medium)

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.

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Example:
Input: [1,2,3,null,5,null,4]
Output: [1, 3, 4]
Explanation:

1 <---
/ \
2 3 <---
\ \
5 4 <---

递归:

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class Solution:
def rightSideView(self, root: TreeNode) -> List[int]:
if not root:
return []
ans = []
def helper(root, depth):
if not root:
return
if depth > len(ans):
ans.append(root.val)
helper(root.right, depth + 1)
helper(root.left, depth + 1)
helper(root, 1)
return ans

非递归:

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class Solution:
def rightSideView(self, root: TreeNode) -> List[int]:
if not root:
return []
ans = []
q = [root]
while q:
ans.append(q[-1].val)
nq = []
for root in q:
if root.left:
nq.append(root.left)
if root.right:
nq.append(root.right)
q = nq
return ans

五刷:面经, 维萨

235. Lowest Common Ancestor of a Binary Search Tree (Easy)

Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.
According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”
Given binary search tree: root = [6,2,8,0,4,7,9,null,null,3,5]
LCA of BST example
Example 1:
Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
Output: 6
Explanation: The LCA of nodes 2 and 8 is 6.

Example 2:
Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
Output: 2
Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

Note:
All of the nodes’ values will be unique.
p and q are different and both values will exist in the BST.

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class Solution:
def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
if not root:
return
if p.val > q.val:
p, q = q, p
if root.val < p.val:
return self.lowestCommonAncestor(root.right, p, q)
if root.val > q.val:
return self.lowestCommonAncestor(root.left, p, q)
return root

解法2:

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class Solution:
def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
if root.val < p.val and root.val < q.val:
return self.lowestCommonAncestor(root.right, p, q)
if root.val > p.val and root.val > q.val:
return self.lowestCommonAncestor(root.left, p, q)
return root

八刷:注意要return f(),否则会return最顶上的root

543. Diameter of Binary Tree (Easy)

Given a binary tree, you need to compute the length of the diameter of the tree. The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass through the root.

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Example:
Given a binary tree
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/ \
2 3
/ \
4 5
Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3].

Note: The length of path between two nodes is represented by the number of edges between them.

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class Solution:
def diameterOfBinaryTree(self, root: TreeNode) -> int:
if not root:
return 0
ans = 0
def helper(root):
nonlocal ans
if not root:
return 0
l = helper(root.left)
r = helper(root.right)
ans = max(ans, l + r)
return max(l, r) + 1
helper(root)
return ans

五刷: 重点是理解题的本质是求左右子树的maxDepth,O(n) 每一步都算一下当前l+r是否大于ans

563. Binary Tree Tilt (Easy)

Given a binary tree, return the tilt of the whole tree.
The tilt of a tree node is defined as the absolute difference between the sum of all left subtree node values and the sum of all right subtree node values. Null node has tilt 0.
The tilt of the whole tree is defined as the sum of all nodes’ tilt.

Example:

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Input:
1
/ \
2 3
Output: 1

Explanation:
Tilt of node 2 : 0
Tilt of node 3 : 0
Tilt of node 1 : |2-3| = 1
Tilt of binary tree : 0 + 0 + 1 = 1
Note:
The sum of node values in any subtree won’t exceed the range of 32-bit integer.
All the tilt values won’t exceed the range of 32-bit integer.

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class Solution:
def findTilt(self, root: TreeNode) -> int:
if not root:
return 0
ans = 0
def helper(root):
nonlocal ans
if not root:
return 0
l = helper(root.left)
r = helper(root.right)
ans += abs(l - r)
return l + r + root.val
helper(root)
return ans

六刷:O(n)

669. Trim a Binary Search Tree (Easy)

Given a binary search tree and the lowest and highest boundaries as L and R, trim the tree so that all its elements lies in [L, R] (R >= L). You might need to change the root of the tree, so the result should return the new root of the trimmed binary search tree.

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Example 1:
Input:
1
/ \
0 2
L = 1
R = 2
Output:
1
\
2

Example 2:
Input:
3
/ \
0 4
\
2
/
1
L = 1
R = 3
Output:
3
/
2
/
1

写法1:

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class Solution:
def trimBST(self, root: TreeNode, L: int, R: int) -> TreeNode:
if not root:
return
l = self.trimBST(root.left, L, R)
r = self.trimBST(root.right, L, R)
if root.val < L:
return r
if root.val > R:
return l
root.left = l
root.right = r
return root

写法2:

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class Solution:
def trimBST(self, root: TreeNode, L: int, R: int) -> TreeNode:
if not root:
return
if root.val < L:
return self.trimBST(root.right, L, R)
if root.val > R:
return self.trimBST(root.left, L, R)
root.left = self.trimBST(root.left, L, R)
root.right = self.trimBST(root.right, L, R)
return root

七刷:每层三种情况:1:小于最小,则返回右子树传上来的root;2:大于最大,则返回左子树传上来的root;3.大小之间,递归

687. Longest Univalue Path (Easy)

again!
Given a binary tree, find the length of the longest path where each node in the path has the same value. This path may or may not pass through the root.
The length of path between two nodes is represented by the number of edges between them.

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Example 1:
Input:
5
/ \
4 5
/ \ \
1 1 5
Output: 2

Example 2:
Input:
1
/ \
4 5
/ \ \
4 4 5
Output: 2

Note: The given binary tree has not more than 10000 nodes. The height of the tree is not more than 1000.

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class Solution:
def longestUnivaluePath(self, root: TreeNode) -> int:
if not root:
return 0
ans = 0
def helper(root, pre):
nonlocal ans
if not root:
return 0
l = helper(root.left, root)
r = helper(root.right, root)
ans = max(ans, l + r)
if pre and root.val == pre.val:
return max(l, r) + 1
else:
return 0
helper(root, None)
return ans

六刷

114. Flatten Binary Tree to Linked List (Medium)

Given a binary tree, flatten it to a linked list in-place.

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For example, given the following tree:
1
/ \
2 5
/ \ \
3 4 6
The flattened tree should look like:
1
\
2
\
3
\
4
\
5
\
6

解法1:

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class Solution:
def flatten(self, root: TreeNode) -> None:
if not root:
return
l = self.flatten(root.left)
r = self.flatten(root.right)
if l:
root.right = l
root.left = None
while l.right:
l = l.right
l.right = r
return root

九刷:高频

236. Lowest Common Ancestor of a Binary Tree (Medium)

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”
Given the following binary tree: root = [3,5,1,6,2,0,8,null,null,7,4]

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7
     _______3______
/ \
___5__ ___1__
/ \ / \
6 _2 0 8
/ \
7 4

Example 1:
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of of nodes 5 and 1 is 3.

Example 2:
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself
according to the LCA definition.

Note:
All of the nodes’ values will be unique.
p and q are different and both values will exist in the binary tree.

递归:

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class Solution:
def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
if not root:
return
if root == p or root == q:
return root
l = self.lowestCommonAncestor(root.left, p, q)
r = self.lowestCommonAncestor(root.right, p, q)
if l and not r:
return l
if not l and r:
return r
if l and r:
return root

非递归:

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class Solution:
def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
parent = {root: None}
queue = [root]
while p not in parent or q not in parent:
root = queue.pop(0)
if root.left:
parent[root.left] = root
queue.append(root.left)
if root.right:
parent[root.right] = root
queue.append(root.right)
ancestors = [p]
while p:
p = parent[p]
ancestors.append(p)
while q not in ancestors:
q = parent[q]
return q

12刷:递归:注意最后返回root需要if l and r这个条件。非递归:bfs遍历,建立parent关系,p存入ancestors set,遍历p的parent入ancestors,遍历q和q的parent,在ancestors中即返回。 注意queue写全了,不然会与q节点起名冲突

285. Inorder Successor in BST (Medium) 带锁

LinC 448. Inorder Successor in BST (Medium)
Given a binary search tree and a node in it, find the in-order successor of that node in the BST.
If the given node has no in-order successor in the tree, return null.
It’s guaranteed p is one node in the given tree. (You can directly compare the memory address to find p)

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Example
Given tree = [2,1] and node = 1:
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/
1
return node 2.

Given tree = [2,1,3] and node = 2:
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/ \
1 3
return node 3.

Challenge
O(h), where h is the height of the BST.

O(n):

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class Solution:
def inorderSuccessor(self, root, p):
pre = None
ans = None
def helper(root):
nonlocal pre, ans
if not root:
return
helper(root.left)
if pre == p:
ans = root
pre = root
helper(root.right)
helper(root)
return ans

非递归:

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class Solution:
def inorderSuccessor(self, root, p):
pre = None
stack = []
while root or stack:
if root:
stack.append(root)
root = root.left
else:
root = stack.pop()
if pre == p:
return root
pre = root
root = root.right

O(h)

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class Solution:
def inorderSuccessor(self, root, p):
if not root:
return
suc = None
while root != p:
if root.val < p.val:
root = root.right
else:
suc = root
root = root.left
if not root.right:
return suc
root = root.right
while root.left:
root = root.left
return root

八刷:递归O(n)解法:注意找到ans以后如果想return需要将pre = None,否则会锁定pre,导致上层的 pre == p,而将上层的root赋值到ans中。
O(h)解法:找p的successor,那么如果root比p小p在右子树,反之p在左子树,去左子树之前要记一下suc节点,以防找到的root.val == p.val节点无右节点(即需要返回上层的suc)。如有右节点,则返回右节点中最小的(往右走一个然后返回最左节点)

98. Validate Binary Search Tree (Medium)

Given a binary tree, determine if it is a valid binary search tree (BST).
Assume a BST is defined as follows:
The left subtree of a node contains only nodes with keys less than the node’s key.
The right subtree of a node contains only nodes with keys greater than the node’s key.
Both the left and right subtrees must also be binary search trees.

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Example 1:
Input:
2
/ \
1 3
Output: true

Example 2:

5
/ \
1 4
/ \
3 6
Output: false

Explanation: The input is: [5,1,4,null,null,3,6]. The root node's value
is 5 but its right child's value is 4.

高频
递归:

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class Solution:
def isValidBST(self, root: TreeNode) -> bool:
def helper(root, minV, maxV):
if not root:
return True
if root.val <= minV or root.val >= maxV:
return False
return helper(root.left, minV, root.val) and helper(root.right, root.val, maxV)
return helper(root, -sys.maxsize, sys.maxsize)

inorder 中序遍历非递归无额外数组:

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class Solution:
def isValidBST(self, root):
stack = []
pre = None
while root or stack:
if root:
stack.append(root)
root = root.left
else:
root = stack.pop(-1)
if pre and pre.val >= root.val:
return False
pre = root
root = root.right
return True

inorder 中序遍历递归无额外数组:

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class Solution:
def isValidBST(self, root: TreeNode) -> bool:
pre = None
ans = True
def inorder(root):
nonlocal pre, ans
if not root:
return
inorder(root.left)
if pre and pre.val >= root.val:
ans = False
return
pre = root
inorder(root.right)
inorder(root)
return ans

inorder 中序遍历递归额外数组:

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class Solution:
def isValidBST(self, root):
res = []
self.inOrder(root, res)
for i in range(1, len(res)):
if res[i] <= res[i - 1]:
return False
return True
def inOrder(self, root, res):
if not root:
return
self.inOrder(root.left, res)
res.append(root.val)
self.inOrder(root.right, res)

高频,七刷:五种解法:1.利用BST性质的递归,2,3.中序遍历递归(有/无额外数组),4,5.中序遍历非递归(有/无额外数组)
中序遍历非递归无额外数组:…if pre and n.val <= pre.val: return False; pre = n…
BST性质递归:…if root.val <= minV or root.val >= maxV…

572. Subtree of Another Tree (Easy)

Given two non-empty binary trees s and t, check whether tree t has exactly the same structure and node values with a subtree of s. A subtree of s is a tree consists of a node in s and all of this node’s descendants. The tree s could also be considered as a subtree of itself.

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Example 1:
Given tree s:
3
/ \
4 5
/ \
1 2
Given tree t:
4
/ \
1 2
Return true, because t has the same structure and node values with a subtree of s.

Example 2:
Given tree s:

3
/ \
4 5
/ \
1 2
/
0
Given tree t:
4
/ \
1 2
Return false.

again!

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class Solution:
def isSubtree(self, s: TreeNode, t: TreeNode) -> bool:
if not s and t:
return False
def helper(s, t):
if not s and not t:
return True
if (s and not t) or (not s and t) or (s.val != t.val):
return False
return helper(s.left, t.left) and helper(s.right, t.right)
return helper(s, t) or self.isSubtree(s.left, t) or self.isSubtree(s.right, t)

七刷:注意:要判断s是否为空,否则递归入口会因为s为空出错;既有内层helper的递归,又有外层isSubtree递归遍历s树

606. Construct String from Binary Tree (Easy)

You need to construct a string consists of parenthesis and integers from a binary tree with the preorder traversing way.
The null node needs to be represented by empty parenthesis pair “()”. And you need to omit all the empty parenthesis pairs that don’t affect the one-to-one mapping relationship between the string and the original binary tree.

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Example 1:
Input: Binary tree: [1,2,3,4]
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/ \
2 3
/
4
Output: "1(2(4))(3)"
Explanation: Originallay it needs to be "1(2(4)())(3()())",
but you need to omit all the unnecessary empty parenthesis pairs.
And it will be "1(2(4))(3)".

Example 2:
Input: Binary tree: [1,2,3,null,4]
1
/ \
2 3
\
4

Output: "1(2()(4))(3)"
Explanation: Almost the same as the first example,
except we can't omit the first parenthesis pair to break the one-to-one mapping relationship between the input and the output.

写法1:

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class Solution:
def tree2str(self, t: TreeNode) -> str:
if not t:
return ''
l = self.tree2str(t.left)
r = self.tree2str(t.right)
if l and r:
return f"{t.val}({l})({r})"
if l and not r:
return f"{t.val}({l})"
if not l and r:
return f"{t.val}()({r})"
return f"{t.val}"

写法2:

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class Solution:
def tree2str(self, t: TreeNode) -> str:
if not t:
return ''
ans = ''
def helper(root):
nonlocal ans
ans += str(root.val)
if root.left:
ans += '('
helper(root.left)
ans += ')'
elif root.right:
ans += '()'
if root.right:
ans += '('
helper(root.right)
ans += ')'
helper(t)
return ans

六刷

297. Serialize and Deserialize Binary Tree (Hard)

Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.
Design an algorithm to serialize and deserialize a binary tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary tree can be serialized to a string and this string can be deserialized to the original tree structure.

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Example:
You may serialize the following tree:
1
/ \
2 3
/ \
4 5
as "[1,2,3,null,null,4,5]"

Clarification: The above format is the same as how LeetCode serializes a binary tree. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.
Note: Do not use class member/global/static variables to store states. Your serialize and deserialize algorithms should be stateless.

递归:

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class Codec:
def serialize(self, root):
if not root:
return
res = []
def helper(root):
if not root:
res.append('#')
return
res.append(str(root.val))
helper(root.left)
helper(root.right)
helper(root)
return ','.join(res)

def deserialize(self, data):
if not data:
return
arr = data.split(',')
def helper():
rootV = arr.pop(0)
if rootV == '#':
return
root = TreeNode(int(rootV))
root.left = helper()
root.right = helper()
return root
return helper()

非递归:

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class Codec:
def serialize(self, root) -> str:
if not root:
return
q = [root]
res = []
while q:
root = q.pop(0)
if not root:
res.append('#')
else:
res.append(str(root.val))
q.append(root.left)
q.append(root.right)
return ','.join(res)

def deserialize(self, data: str) -> TreeNode:
if not data:
return
arr = data.split(',')
root = TreeNode(int(arr.pop(0)))
nq = [root]
while arr:
lV, rV = arr.pop(0), arr.pop(0)
node = nq.pop(0)
if lV != '#':
node.left = TreeNode(int(lV))
nq.append(node.left)
if rV != '#':
node.right = TreeNode(int(rV))
nq.append(node.right)
return root

六刷:要牢记递归和非递归产生的结果是不一样的。递归serialize的结果是:”1,2,#,#,3,4,#,#,5,#,#”,非递归的seriallize结果是:“1,2,3,#,#,4,5,#,#,#,#”

536. Construct Binary Tree from String (Medium) 带锁

LinC 880. Construct Binary Tree from String (Medium)
You need to construct a binary tree from a string consisting of parenthesis and integers.
The whole input represents a binary tree. It contains an integer followed by zero, one or two pairs of parenthesis. The integer represents the root’s value and a pair of parenthesis contains a child binary tree with the same structure.
You always start to construct the left child node of the parent first if it exists.

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Example 1:
Input: "-4(2(3)(1))(6(5))"
Output: {-4,2,6,3,1,5}
Explanation:
The output is look like this:
-4
/ \
2 6
/ \ /
3 1 5

Example 2:
Input: "1(-1)"
Output: {1,-1}
Explanation:
The output is look like this:
1
/
-1

Notice
There will only be ‘(‘, ‘)’, ‘-‘ and ‘0’ ~ ‘9’ in the input string.
An empty tree is represented by “” instead of “()”.

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class Solution:
def str2tree(self, s):
if not s:
return
stack = []
i = 0
while i < len(s):
cur = ''
while s[i] and s[i] not in '()':
cur += s[i]
i += 1
if cur:
root = TreeNode(int(cur))
if stack:
if stack[-1].left:
stack[-1].right = root
else:
stack[-1].left = root
stack.append(root)
if s[i] == ")":
stack.pop()
i += 1
return stack[0]

七刷:Facebook tag,递归的方法不适合面试时写,将此较好想的方法写熟。注意:string没有pop()

105. Construct Binary Tree from Preorder and Inorder Traversal (Medium)

Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.

For example, given
preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]
Return the following binary tree:

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5
  3
/ \
9 20
/ \
15 7

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class Solution:
def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode:
if not preorder:
return
root = TreeNode(preorder[0])
lLen = inorder.index(preorder[0])
root.left = self.buildTree(preorder[1:lLen + 1], inorder[:lLen])
root.right = self.buildTree(preorder[lLen + 1:], inorder[lLen + 1:])
return root

O(n)时间和空间:

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class Solution:
def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode:
if not preorder:
return
iMap = {}
for i, v in enumerate(inorder):
iMap[v] = i
def helper(preS, preE, inS, inE):
if preS > preE:
return
root = TreeNode(preorder[preS])
if preS < preE:
lEIIn = iMap[preorder[preS]] - 1
lEIPre = preS + 1 + lEIIn - inS
root.left = helper(preS + 1, lEIPre, inS, lEIIn)
root.right = helper(lEIPre + 1, preE, lEIIn + 2, inE)
return root
return helper(0, len(preorder) - 1, 0, len(inorder) - 1)

六刷:高频

106. Construct Binary Tree from Inorder and Postorder Traversal (Medium)

Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.

For example, given
inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]
Return the following binary tree:

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4
5
  3
/ \
9 20
/ \
15 7

O(n^2) time and space:

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class Solution:
def buildTree(self, inorder: List[int], postorder: List[int]) -> TreeNode:
if not postorder:
return
root = TreeNode(postorder[-1])
lLen = inorder.index(postorder[-1])
root.left = self.buildTree(inorder[:lLen], postorder[:lLen])
root.right = self.buildTree(inorder[lLen + 1:], postorder[lLen:-1])
return root

O(n) time and space:

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class Solution:
def buildTree(self, inorder: List[int], postorder: List[int]) -> TreeNode:
if not postorder:
return
iMap = {}
for i, v in enumerate(inorder):
iMap[v] = i
def helper(inS, inE, postS, postE):
if inS > inE:
return
root = TreeNode(postorder[postE])
ilE = iMap[root.val] - 1
plE = postS + ilE - inS
root.left = helper(inS, ilE, postS, plE)
root.right = helper(ilE + 2, inE, plE + 1, postE - 1)
return root
return helper(0, len(inorder) - 1, 0, len(postorder) - 1)

七刷: 高频

889. Construct Binary Tree from Preorder and Postorder Traversal (Medium)

Return any binary tree that matches the given preorder and postorder traversals.
Values in the traversals pre and post are distinct positive integers.

Example 1:
Input: pre = [1,2,4,5,3,6,7], post = [4,5,2,6,7,3,1]
Output: [1,2,3,4,5,6,7]

Note:
1 <= pre.length == post.length <= 30
pre[] and post[] are both permutations of 1, 2, …, pre.length.
It is guaranteed an answer exists. If there exists multiple answers, you can return any of them.

O(n)时间和空间解:

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class Solution:
def constructFromPrePost(self, pre: List[int], post: List[int]) -> TreeNode:
if not pre:
return
iMap = {}
for i, v in enumerate(post):
iMap[v] = i
def helper(preS, preE, postS, postE):
if preS > preE:
return
root = TreeNode(pre[preS])
if preS < preE:
lEIPost = iMap[pre[preS + 1]]
lEIPre = preS + 1 + lEIPost - postS
root.left = helper(preS + 1, lEIPre, postS, lEIPost)
root.right = helper(lEIPre + 1, preE, lEIPost + 1, postE - 1)
return root
return helper(0, len(pre) - 1, 0, len(post) - 1)

六刷:Facebook tag。需要一个insight:root in left subtree of pre show up last in left subtree of post

865. Smallest Subtree with all the Deepest Nodes (Medium)

Given a binary tree rooted at root, the depth of each node is the shortest distance to the root.
A node is deepest if it has the largest depth possible among any node in the entire tree.
The subtree of a node is that node, plus the set of all descendants of that node.
Return the node with the largest depth such that it contains all the deepest nodes in its subtree.

Example 1:
Input: [3,5,1,6,2,0,8,null,null,7,4]
Output: [2,7,4]
Explanation:
Smallest Subtree with all the Deepest Nodes example
We return the node with value 2, colored in yellow in the diagram.
The nodes colored in blue are the deepest nodes of the tree.
The input “[3, 5, 1, 6, 2, 0, 8, null, null, 7, 4]” is a serialization of the given tree.
The output “[2, 7, 4]” is a serialization of the subtree rooted at the node with value 2.
Both the input and output have TreeNode type.

Note:
The number of nodes in the tree will be between 1 and 500.
The values of each node are unique.

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class Solution:
def subtreeWithAllDeepest(self, root: TreeNode) -> TreeNode:
if not root:
return
parents = {}
deepest = []
maxDepth = 0
def helper(root, pre, depth):
nonlocal maxDepth, deepest
if not root:
return
parents[root] = pre
if depth == maxDepth:
deepest.append(root)
elif depth > maxDepth:
deepest = [root]
maxDepth = depth
helper(root.left, root, depth + 1)
helper(root.right, root, depth + 1)
helper(root, None, 1)
while len(deepest) > 1:
deepest = set([parents[root] for root in deepest])
return list(deepest)[0]

六刷:Facebook tag

426. Convert Binary Search Tree to Sorted Doubly Linked List (Medium) 带锁

Linc 1534. Convert Binary Search Tree to Sorted Doubly Linked List

again!

Convert a BST to a sorted circular doubly-linked list in-place. Think of the left and right pointers as synonymous to the previous and next pointers in a doubly-linked list.
Let’s take the following BST as an example, it may help you understand the problem better:
bstdll original bst example
We want to transform this BST into a circular doubly linked list. Each node in a doubly linked list has a predecessor and successor. For a circular doubly linked list, the predecessor of the first element is the last element, and the successor of the last element is the first element.
The figure below shows the circular doubly linked list for the BST above. The “head” symbol means the node it points to is the smallest element of the linked list.
bstdll return dll example
Specifically, we want to do the transformation in place. After the transformation, the left pointer of the tree node should point to its predecessor, and the right pointer should point to its successor. We should return the pointer to the first element of the linked list.
The figure below shows the transformed BST. The solid line indicates the successor relationship, while the dashed line means the predecessor relationship.
bstdll return bst example

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Example 1:
Input: {4,2,5,1,3}
4
/ \
2 5
/ \
1 3
Output: "left:1->5->4->3->2 right:1->2->3->4->5"
Explanation:
Left: reverse output
Right: positive sequence output

Example 2:
Input: {2,1,3}
2
/ \
1 3
Output: "left:1->3->2 right:1->2->3"

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"""
Definition of TreeNode:
class TreeNode:
def __init__(self, val):
self.val = val
self.left, self.right = None, None
"""
class Solution:
def treeToDoublyList(self, root):
if not root:
return
head = None
pre = None
def helper(root):
nonlocal head, pre
if not root:
return
helper(root.left)
if not head:
head = root
if pre:
root.left = pre
pre.right = root
pre = root
helper(root.right)
helper(root)
head.left = pre
pre.right = head
return head

四刷:Facebook tag。中序遍历可以升序遍历。连接相邻结点,需要用变量 pre 记录上一个遍历到的结点。需要变量head来指向最小(最左)的节点。在递归函数中,先判空,之后对左子结点递归调用,一直递归到最左结点。此时如果 head 为空的话,那么当前就是最左结点,赋值给 head 然后给 pre,对于之后遍历到的结点,就可以和 pre 接上

298. Binary Tree Longest Consecutive Sequence (Medium) 带锁

LinC 595. Binary Tree Longest Consecutive Sequence
Given a binary tree, find the length of the longest consecutive sequence path.
The path refers to any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The longest consecutive path need to be from parent to child (cannot be the reverse).

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Example
Example 1:
Input:
1
\
3
/ \
2 4
\
5
Output:3
Explanation:
Longest consecutive sequence path is 3-4-5, so return 3.

Example 2:
Input:
2
\
3
/
2
/
1
Output:2
Explanation:
Longest consecutive sequence path is 2-3,not 3-2-1, so return 2.

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class Solution:
def longestConsecutive(self, root):
if not root:
return 0
ans = 0
def helper(root, pre, cur):
nonlocal ans
if not root:
return
if root.val == pre.val + 1:
cur += 1
else:
cur = 1
ans = max(ans, cur)
helper(root.left, root, cur)
helper(root.right, root, cur)
helper(root, root, 1)
return ans

非递归:

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class Solution:
def longestConsecutive(self, root):
if not root:
return 0
ans = 0
q = [(root, root, 1)]
while q:
root, pre, cur = q.pop(0)
if not root:
continue
if root.val == pre.val + 1:
cur += 1
else:
cur = 1
ans = max(ans, cur)
q.append((root.left, root, cur))
q.append((root.right, root, cur))
return ans

五刷:Facebook tag

897. Increasing Order Search Tree (Easy)

Given a binary search tree, rearrange the tree in in-order so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only 1 right child.

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Example 1:
Input: [5,3,6,2,4,null,8,1,null,null,null,7,9]
5
/ \
3 6
/ \ \
2 4 8
/ / \
1 7 9
Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]
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2
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3
\
4
\
5
\
6
\
7
\
8
\
9

Note:
The number of nodes in the given tree will be between 1 and 100.
Each node will have a unique integer value from 0 to 1000.

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class Solution:
def increasingBST(self, root: TreeNode) -> TreeNode:
dummy = pre = TreeNode(0)
def helper(root):
nonlocal pre
if not root:
return
helper(root.left)
root.left = None
pre.right = root
pre = root
helper(root.right)
helper(root)
return dummy.right

三刷:Facebook tag,关键要建立pre

Tree based BFS 基于树的 BFS

102. Binary Tree Level Order Traversal (Medium)

Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).

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For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]

递归:

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class Solution:
def levelOrder(self, root: TreeNode) -> List[List[int]]:
if not root:
return []
ans = []
def helper(root, depth):
if not root:
return
if depth == len(ans):
ans.append([root.val])
else:
ans[depth].append(root.val)
helper(root.left, depth + 1)
helper(root.right, depth + 1)
helper(root, 0)
return ans

非递归:

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class Solution:
def levelOrder(self, root: TreeNode) -> List[List[int]]:
if not root:
return []
ans = []
q = [root]
while q:
cur = []
for _ in range(len(q)):
n = q.pop(0)
cur.append(n.val)
if n.left:
q.append(n.left)
if n.right:
q.append(n.right)
ans.append(cur)
return ans

四刷:高频

107. Binary Tree Level Order Traversal II (Easy)

Given a binary tree, return the bottom-up level order traversal of its nodes’ values. (ie, from left to right, level by level from leaf to root).

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For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[
[15,7],
[9,20],
[3]
]

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class Solution:
def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
if not root:
return []
ans = []
def helper(root, depth):
if not root:
return
if depth == len(ans) :
ans.insert(0, [])
ans[len(ans) - depth - 1].append(root.val)
helper(root.left, depth + 1)
helper(root.right, depth + 1)
helper(root, 0)
return ans

非递归:

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class Solution:
def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
if not root:
return []
q = [root]
ans = []
while q:
nq = []
cur = []
for root in q:
cur.append(root.val)
if root.left:
nq.append(root.left)
if root.right:
nq.append(root.right)
ans.insert(0, cur)
q = nq
return ans

四刷:高频。list.insert(0, x), 也可用deque的appendleft

103. Binary Tree Zigzag Level Order Traversal (Medium)

LinC 71. Binary Tree Zigzag Order Traversal
Given a binary tree, return the zigzag level order traversal of its nodes’ values. (ie, from left to right, then right to left for the next level and alternate between).

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For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its zigzag level order traversal as:
[
[3],
[20,9],
[15,7]
]

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class Solution:
def zigzagLevelOrder(self, root):
if not root:
return []
q = [root]
ans = []
zig = True
while q:
ans.append([root.val for root in q])
if not zig:
ans[-1].reverse()
zig = not zig
nq = []
for root in q:
if root.left:
nq.append(root.left)
if root.right:
nq.append(root.right)
q = nq
return ans

二刷:高频,递归的话可以用list.insert(0, val)来处理偶数层的reverse

958. Check Completeness of a Binary Tree (Medium)

Given a binary tree, determine if it is a complete binary tree.
Definition of a complete binary tree from Wikipedia:
In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between 1 and 2h nodes inclusive at the last level h.

Example 1:
Check Completeness of a Binary Tree example1
Input: [1,2,3,4,5,6]
Output: true
Explanation: Every level before the last is full (ie. levels with node-values {1} and {2, 3}), and all nodes in the last level ({4, 5, 6}) are as far left as possible.

Example 2:
Check Completeness of a Binary Tree example1
Input: [1,2,3,4,5,null,7]
Output: false
Explanation: The node with value 7 isn’t as far left as possible.

Note:
The tree will have between 1 and 100 nodes.

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class Solution:
def isCompleteTree(self, root: TreeNode) -> bool:
q = [root]
noMore = False
while q:
root = q.pop(0)
if not root:
noMore = True
continue
if root:
if noMore:
return False
else:
q.append(root.left)
q.append(root.right)
return True

二刷

Binary Search & LogN Algorithm

二分法模板: start + 1 < end; start + (end - start) / 2; A[mid] ==, <, >; A[start] A[end] ? target

704. Binary Search (Easy)

lintcode’s version

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Find any position of a target number in a sorted array. Return -1 if target does not exist.

Example
Given [1, 2, 2, 4, 5, 5].

For target = 2, return 1 or 2.

For target = 5, return 4 or 5.

For target = 6, return -1.

Challenge
O(logn) time

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class Solution:
"""
@param: nums: An integer array sorted in ascending order
@param: target: An integer
@return: An integer
"""
def findPosition(self, nums, target):
# write your code here
if (len(nums) == 0):
return -1
start, end = 0, len(nums) - 1
while (start + 1 < end):
mid = start + (end - start) // 2
if (nums[mid] == target):
return mid
elif (nums[mid] < target):
start = mid
else:
end = mid
if (nums[start] == target):
return start
if (nums[end] == target):
return end
return -1

总结:背好模板,lintcode 的 test case 包含空输入数组,需要 python3 的 // 整除运算符才能过

二刷:

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class Solution:
def search(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
if len(nums) == 0 or (len(nums) == 1 and nums[0] != target):
return -1
return self.helper(nums, target, 0, len(nums) - 1)
def helper(self, nums, target, start, end):
if (start > end):
return -1
if (start + 1 == end):
if nums[end] == target:
return end
if nums[start] == target:
return start
else:
return -1
mid = start + (end - start) // 2
if (nums[mid] == target):
return mid
elif (nums[mid] < target):
start = mid
else:
end = mid
return self.helper(nums, target, start, end)

总结:不背模板也能写。 但是写出来不如模板的优雅。如果递归调用前面不加 return 的话,还会发生不 return 的情况

LinC 14. First Position of Target (Easy)

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Description
For a given sorted array (ascending order) and a target number, find the first index of this number in O(log n) time complexity.

If the target number does not exist in the array, return -1.

Have you met this question in a real interview?
Example
If the array is [1, 2, 3, 3, 4, 5, 10], for given target 3, return 2.

Challenge
If the count of numbers is bigger than 2^32, can your code work properly?

思路:找到了不要 return,扔掉大的一半,继续找

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class Solution:
"""
@param nums: The integer array.
@param target: Target to find.
@return: The first position of target. Position starts from 0.
"""
def binarySearch(self, nums, target):
# write your code here
if (len(nums) == 0):
return -1
start, end = 0, len(nums) - 1
while (start + 1 < end):
mid = start + (end - start) // 2
if (nums[mid] >= target):
end = mid
else:
start = mid
if (nums[start] == target):
return start
if (nums[end] == target):
return end
return -1

总结:背好模板,模板 v5

278. First Bad Version (Easy)

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You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions after a bad version are also bad.

Suppose you have n versions [1, 2, ..., n] and you want to find out the first bad one, which causes all the following ones to be bad.

You are given an API bool isBadVersion(version) which will return whether version is bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.

Example:

Given n = 5, and version = 4 is the first bad version.

call isBadVersion(3) -> false
call isBadVersion(5) -> true
call isBadVersion(4) -> true

Then 4 is the first bad version.

思路:前面 first position of target 的变体,可以不做

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# The isBadVersion API is already defined for you.
# @param version, an integer
# @return a bool
# def isBadVersion(version):

class Solution(object):
def firstBadVersion(self, n):
"""
:type n: int
:rtype: int
"""
start, end = 0, n
while (start + 1 < end):
mid = start + (end - start) // 2
if (isBadVersion(mid)):
end = mid
else:
start = mid
if (isBadVersion(start)):
return start
if (isBadVersion(end)):
return end
return -1

总结:可不做

二刷:

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# The isBadVersion API is already defined for you.
# @param version, an integer
# @return a bool
# def isBadVersion(version):

class Solution:
def firstBadVersion(self, n):
"""
:type n: int
:rtype: int
"""
if n == 1:
return 1 if isBadVersion(1) else -1
return self.helper(1, n)
def helper(self, start, end):
if start + 1 == end:
return start if isBadVersion(start) else end
mid = start + (end - start) // 2
if isBadVersion(mid):
end = mid
else:
start = mid
return self.helper(start, end)

总结:递归思维更自然。二刷之前没有看模板。模板大法更普适

34. Find First and Last Position of Element in Sorted Array (Medium)

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Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]
Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

思路:二分法找 Target, 两次二分法,一次找左边界,一次找右边界

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class Solution(object):
def searchRange(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
ans = [-1, -1]
if len(nums) == 0:
return ans
l, r = 0, len(nums) - 1
# 找左边界
while l + 1 < r:
mid = l + (r - l ) // 2
if nums[mid] < target:
l = mid
else:
r = mid

if nums[l] == target:
ans[0] = l
elif nums[r] == target:
ans[0] = r
else:
return ans

# 找右边界
r = len(nums) - 1
while l + 1 < r:
mid = l + (r - l) // 2
if nums[mid] <= target:
l = mid
else:
r = mid
if nums[r] == target:
ans[1] = r
elif nums[l] == target:
ans[1] = l
return ans

总结:按今天的水平,写的时候注意 while 的终止条件是 while l + 1 < r (l, r 不要重合就终止循环)。 两年多前写了稍微更简洁些的版本。可以回头再看看能不能写得出。

二刷:

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class Solution:
def searchRange(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""

if len(nums) == 0:
return [-1, -1]

if len(nums) == 1:
return [0, 0] if nums[0] == target else [-1, -1]

start, end = 0, len(nums) - 1
while (start + 1 < end):
mid = start + (end - start) // 2
if nums[mid] == target:
end = mid
if nums[mid] < target:
start = mid
if nums[mid] > target:
end = mid
res1 = end if nums[end] == target else -1
res1 = start if nums[start] == target else res1
if res1 == -1:
return [-1, -1]

start, end = res1, len(nums) - 1
while(start + 1 < end):
mid = start + (end - start) // 2
if nums[mid] == target:
start = mid
if nums[mid] < target:
start = mid
if nums[mid] > target:
end = mid
res2 = start if nums[start] == target else -1
res2 = end if nums[end] == target else res2
if res2 == -1:
return [res1, res1]
else:
return [res1, res2]

总结: 凭借模板大法战胜 100% 的 python 选手。
高频:…if nums[mid] == target: r = mid…res1 = r … res1 = l …else res1; if res1 == -1: return [-1, -1]…res2 = l … res2 = r …else res2; if …: return [res1, res1]…

LinC 61. Search for a Range (Medium)

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Given a sorted array of n integers, find the starting and ending position of a given target value.

If the target is not found in the array, return [-1, -1].

Example
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

Challenge
O(log n) time.

思路:找一个数的第一次和最后一次出现的 index

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class Solution:
"""
@param A: an integer sorted array
@param target: an integer to be inserted
@return: a list of length 2, [index1, index2]
"""
def searchRange(self, A, target):
# write your code here
firstO, lastO = -1, -1
if len(A) == 0:
return [firstO, lastO]
start, end = 0, len(A) - 1
while (start + 1 < end):
mid = start + (end - start) // 2
if (A[mid] < target):
start = mid
else:
end = mid
if (A[end] == target):
firstO = end
if (A[start] == target):
firstO = start
start, end = 0, len(A) - 1
while (start + 1 < end):
mid = start + (end - start) // 2
if (A[mid] <= target):
start = mid
else:
end = mid
if (A[start] == target):
lastO = start
if (A[end] == target):
lastO = end
return [firstO, lastO]

总结:注意检查空输入!

852. Peak Index in a Mountain Array

LinC 585. Maximum Number in Mountain Sequence (Medium)
Given a mountain sequence of n integers which increase firstly and then decrease, find the mountain top.
Example
Given nums = [1, 2, 4, 8, 6, 3] return 8
Given nums = [10, 9, 8, 7], return 10

思路:切一刀,判断递增就扔左边,递减就扔右边, 不然就找到了中点
二刷:

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class Solution:
def peakIndexInMountainArray(self, A):
"""
:type A: List[int]
:rtype: int
"""
start, end = 0, len(A) - 1
while start + 1 < end:
mid = start + (end - start) // 2
if A[mid - 1] < A[mid] < A[mid + 1]:
start = mid
elif A[mid - 1] > A[mid] > A[mid + 1]:
end = mid
else:
return mid
return start if A[start] > A[end] else end

总结:二刷写法跟一刷一样,哪怕是简单的题,题要看清楚, mid min 不要拼错

162. Find Peak Element (Medium)

A peak element is an element that is greater than its neighbors.
Given an input array nums, where nums[i] ≠ nums[i+1], find a peak element and return its index.
The array may contain multiple peaks, in that case return the index to any one of the peaks is fine.
You may imagine that nums[-1] = nums[n] = -∞.

Example 1:
Input: nums = [1,2,3,1]
Output: 2
Explanation: 3 is a peak element and your function should return the index number 2.

Example 2:
Input: nums = [1,2,1,3,5,6,4]
Output: 1 or 5
Explanation: Your function can return either index number 1 where the peak element is 2,
or index number 5 where the peak element is 6.
Note:
Your solution should be in logarithmic complexity.

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class Solution:
def findPeakElement(self, nums: List[int]) -> int:
l, r = 0, len(nums) - 1
while l + 1 < r:
mid = l + (r - l) // 2
if nums[mid - 1] < nums[mid] < nums[mid + 1]:
l = mid
elif nums[mid - 1] < nums[mid] > nums[mid + 1]:
return mid
else:
r = mid
return r if nums[l] < nums[r] else l

面经:Quora。关键要知道切中点,如果是///向上,则顶点在右,如果/^\则找到顶点,否则顶点在左

74. Search a 2D Matrix (Medium)

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Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

Integers in each row are sorted from left to right.
The first integer of each row is greater than the last integer of the previous row.
Example 1:

Input:
matrix = [
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
target = 3
Output: true
Example 2:

Input:
matrix = [
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
target = 13
Output: false

二刷:

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class Solution:
def searchMatrix(self, matrix, target):
"""
:type matrix: List[List[int]]
:type target: int
:rtype: bool
"""
if len(matrix) == 0 or len(matrix[0]) == 0:
return False
rows, cols = len(matrix) - 1, len(matrix[0]) - 1
startR, endR = 0, rows
while startR + 1 < endR:
midR = startR + (endR - startR) // 2
if matrix[midR][0] < target:
startR = midR
elif matrix[midR][0] > target:
endR = midR
else:
return True
if startR < endR:
if matrix[startR][cols] < target:
startR = endR
startC, endC = 0, cols
while startC + 1 < endC:
midC = startC + (endC - startC) // 2
if matrix[startR][midC] < target:
startC = midC
elif matrix[startR][midC] > target:
endC = midC
else:
return True
return False if matrix[startR][startC] != target and matrix[startR][endC] != target else True

高频:

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class Solution:
def searchMatrix(self, matrix, target):
"""
:type matrix: List[List[int]]
:type target: int
:rtype: bool
"""
m = len(matrix)
if m == 0:
return False
n = len(matrix[0])
if n == 0:
return False
rs, re = 0, m - 1
while rs + 1 < re:
rm = rs + (re - rs) // 2
if matrix[rm][0] < target:
rs = rm
elif matrix[rm][0] > target:
re = rm - 1
else:
return True
if matrix[rs][n - 1] < target:
rs = re
return True if target in matrix[rs] else False

1总结:注意检查空输入
2总结:有两种空情况需要判断 [] 和 [[]],击败 100% python3 选手。。。
高频:记住二分查找的模板:…s + 1 < e…。注意判断…if matrix[rs][n - 1] < target:…

153. Find Minimum in Rotated Sorted Array (Medium)

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Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).

Find the minimum element.

You may assume no duplicate exists in the array.

Example 1:
Input: [3,4,5,1,2]
Output: 1

Example 2:
Input: [4,5,6,7,0,1,2]
Output: 0

思路:找 pivot,pivot > 0 时返回 nums[pivot + 1]。找 pivot 时,如果 mid < start, 扔 end, 如果 mid > start 扔 start

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class Solution:
def findMin(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if (nums[0] < nums[len(nums) - 1]):
return nums[0]
start, end = 0, len(nums) - 1
while (start + 1 < end):
mid = start + (end - start) // 2
if (nums[mid] < nums[start]):
end = mid
else:
start = mid
return nums[end]

总结:应改为 Easy 难度的题。
Follow up: 如果有重复的数? 无法保证在 Log(N) 的时间复杂度内解决 例子:[1,1,1,1,1….,1] 里藏着一个 0.最坏情况下需要把每个位置上的1都看一遍,才能找到最后一个有0 的位置. 考点:能想到这个最坏情况的例子

33. Search in Rotated Sorted Array (Medium)

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Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

Your algorithm's runtime complexity must be in the order of O(log n).

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4
Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

思路:第一感觉是得知道 pivot 在哪,有 pivot 一侧不能随便扔,但是更优的方法是查单调的侧是否可以扔

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class Solution:
def search(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
if len(nums) == 0:
return -1
start, end = 0, len(nums) - 1
while (start + 1 < end):
mid = start + (end - start) // 2
if nums[start] < nums[mid]:
if nums[start] <= target <= nums[mid]:
end = mid
else:
start = mid
else:
if nums[mid] <= target <= nums[end]:
start = mid
else:
end = mid
if nums[start] == target:
return start
if nums[end] == target:
return end
return -1

总结: 注意 [1, 3, 5] target 为 1 这种边界条件, 判断 target 在单调这边需要加等号

二刷:

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class Solution:
def search(self, nums: 'List[int]', target: 'int') -> 'int':
if len(nums) == 0:
return -1
if len(nums) == 1:
return 0 if nums[0] == target else -1
start, end = 0, len(nums) - 1
while start + 1 < end:
mid = start + (end - start) // 2
if nums[mid] == target:
return mid
# pivot 在左
elif nums[mid] < nums[start]:
if nums[mid] < target <= nums[end] :
start = mid + 1
else:
end = mid - 1
# piviot 在右
else:
if nums[mid] > target >= nums[start]:
end = mid - 1
else:
start = mid + 1
if nums[start] == target:
return start
if nums[end] == target:
return end
return -1

总结:判断是否保留单调一边不能只看 nums[mid], 扔一侧的时候可以多扔一个 mid + 1 或 mid - 1
高频:…while l + 1 < r…if nums[mid] < nums[l]: if nums[mid] < target <= nums[r]: …

81. Search in Rotated Sorted Array II (Medium)

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Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]).

You are given a target value to search. If found in the array return true, otherwise return false.

Example 1:

Input: nums = [2,5,6,0,0,1,2], target = 0
Output: true
Example 2:

Input: nums = [2,5,6,0,0,1,2], target = 3
Output: false
Follow up:

This is a follow up problem to Search in Rotated Sorted Array, where nums may contain duplicates.
Would this affect the run-time complexity? How and why?
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class Solution:
def search(self, nums: List[int], target: int) -> bool:
if len(nums) == 0:
return False
l, r = 0, len(nums) - 1
while l + 1 < r:
mid = l + (r - l) // 2
if nums[mid] == target:
return True
if nums[l] < nums[mid]:
if nums[l] <= target < nums[mid]:
r = mid - 1
else:
l = mid + 1
elif nums[l] > nums[mid]:
if nums[mid] < target <= nums[r]:
l = mid + 1
else:
r = mid - 1
else:
l += 1
if nums[l] == target or nums[r] == target:
return True
return False

高频:注意r = len(nums) - 1,判断nums[mid] == nums[l] 的情况下l += 1

69. Sqrt(x) (Easy)

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Implement int sqrt(int x).

Compute and return the square root of x, where x is guaranteed to be a non-negative integer.

Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.

Example 1:

Input: 4
Output: 2
Example 2:

Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since
the decimal part is truncated, 2 is returned.
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class Solution:
def mySqrt(self, x: int) -> int:
l, r = 0, x
while l + 1 < r:
mid = l + (r - l) // 2
if mid * mid <= x < (mid + 1) * (mid + 1):
return mid
elif mid * mid < x:
l = mid + 1
else:
r = mid - 1
return l if l * l <= x < (l + 1) * (l + 1) else r

高频:统一模板…l + 1 < r…if mid mid <= x <(mid + 1) (mid + 1)…return l if l * l <= x <…

35. Search Insert Position (Easy)

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Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.

You may assume no duplicates in the array.

Example 1:

Input: [1,3,5,6], 5
Output: 2
Example 2:

Input: [1,3,5,6], 2
Output: 1
Example 3:

Input: [1,3,5,6], 7
Output: 4
Example 4:

Input: [1,3,5,6], 0
Output: 0
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class Solution:
def searchInsert(self, nums: List[int], target: int) -> int:
l, r = 0, len(nums) - 1
while l + 1 < r:
mid = l + (r - l) // 2
if nums[mid] == target:
return mid
elif nums[mid] < target:
l = mid
else:
r = mid
if target <= nums[l]:
return l
if target <= nums[r]:
return r
return len(nums)

高频:…l = mid…if target <= nums[l]: return l…return len(nums)

658. Find K Closest Elements (Medium)

LinC 460. Find K Closest Elements (Medium)

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Given a sorted array, two integers k and x, find the k closest elements to x in the array. The result should also be sorted in ascending order. If there is a tie, the smaller elements are always preferred.

Example 1:
Input: [1,2,3,4,5], k=4, x=3
Output: [1,2,3,4]
Example 2:
Input: [1,2,3,4,5], k=4, x=-1
Output: [1,2,3,4]
Note:
The value k is positive and will always be smaller than the length of the sorted array.
Length of the given array is positive and will not exceed 104
Absolute value of elements in the array and x will not exceed 104
UPDATE (2017/9/19):
The arr parameter had been changed to an array of integers (instead of a list of integers). Please reload the code definition to get the latest changes.

三刷

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class Solution:
def findClosestElements(self, arr: List[int], k: int, x: int) -> List[int]:
n = len(arr)
l, r = 0, n - 1
while l + 1 < r:
mid = l + (r - l) // 2
if arr[mid] < x:
l = mid
else:
r = mid
cnt = 0
res = []
while l >= 0 and r <= n - 1 and cnt < k:
if abs(arr[l] - x) <= abs(arr[r] - x):
res.append(arr[l])
l -= 1
else:
res.append(arr[r])
r += 1
cnt += 1
while l >= 0 and cnt < k:
res.append(arr[l])
l -= 1
cnt += 1
while r <= n - 1 and cnt < k:
res.append(arr[r])
r += 1
cnt += 1
return sorted(res)

三刷:整理一,二刷代码。算法:1.二分法查找 target,将 l, r 指针放到正确的位置;2.左右按 diff 走 k;3.往左走到底,往右走到底…if nums[mid] < x: l = mid else: r = mid
网上还有一种很妖的O(logN)算法,破坏了模板,核心原理是l, r = 0, len(arr) - k…if abs(arr[mid] - x) > abs(arr[mid + k] - x): l = mid + 1 else: r = mid; return arr[l: l + k]

Two pointers

LinC 373. Partition Array by Odd and Even (Easy)

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Partition an integers array into odd number first and even number second.

Example
Given [1, 2, 3, 4], return [1, 3, 2, 4]

思路:双指针一头一尾,碰到不符合的就换。

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class Solution:
"""
@param: nums: an array of integers
@return: nothing
"""
def partitionArray(self, nums):
# write your code here
if len(nums) < 2:
return
l, r = 0, len(nums) - 1
while l < r:
while l < r and nums[l] % 2 != 0:
l += 1
while l < r and nums[r] % 2 == 0:
r -= 1
if nums[l] % 2 == 0 or nums[r] % 2 != 0:
nums[l], nums[r] = nums[r], nums[l]
if nums[l] % 2 == 0 or nums[r] % 2 != 0:
nums[l], nums[r] = nums[r], nums[l]

总结:送两个测试数据进去就能写对。 最后两个 if 可以简化。

26. Remove Duplicates from Sorted Array (Easy)

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Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

Example 1:

Given nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.

It doesn't matter what you leave beyond the returned length.
Example 2:

Given nums = [0,0,1,1,1,2,2,3,3,4],

Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.

It doesn't matter what values are set beyond the returned length.
Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
print(nums[i]);
}

思路:简单题, 慢指针只有在快指针碰到不同的值才走。

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class Solution:
def removeDuplicates(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if len(nums) == 0:
return 0
slow, fast = 0, 1
while fast < len(nums):
if nums[fast] == nums[slow]:
fast += 1
else:
slow += 1
nums[slow] = nums[fast]
fast += 1
return slow + 1

总结:纯热身,秒解

二刷

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class Solution:
def removeDuplicates(self, nums: List[int]) -> int:
if len(nums) <= 1:
return len(nums)
slow, fast = 0, 1
while fast < len(nums):
if nums[slow] == nums[fast]:
fast += 1
else:
slow += 1
nums[slow], nums[fast] = nums[fast], nums[slow]
fast += 1
return slow + 1

总结:虽然是容易热身题,却要思考两个问题,第一,数组需要 in place sort, 需要利用已经排好序这个条件来在 slow 往前一个以后交换 slow 和 fast 的数; 第二,返回 slow + 1 可以省一个 ans 变量
高频: …else: slow += 1; nums[slow], nums[fast] = nums[fast], nums[slow]; fast += 1…

80. Remove Duplicates from Sorted Array II (Medium)

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Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

Example 1:

Given nums = [1,1,1,2,2,3],

Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively.

It doesn't matter what you leave beyond the returned length.
Example 2:

Given nums = [0,0,1,1,1,1,2,3,3],

Your function should return length = 7, with the first seven elements of nums being modified to 0, 0, 1, 1, 2, 3 and 3 respectively.

It doesn't matter what values are set beyond the returned length.
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class Solution:
def removeDuplicates(self, nums: List[int]) -> int:
if not nums:
return 0
w = 0
for i, n in enumerate(nums):
if i < 2 or n != nums[w - 2]:
nums[w] = n
w += 1
return w

高频:反正两周前的代码也看不懂了,抄个简单一点的…if i < 2 or n != nums[w - 2]

28. Implement strStr() (Easy)

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Implement strStr().

Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.

Example 1:

Input: haystack = "hello", needle = "ll"
Output: 2
Example 2:

Input: haystack = "aaaaa", needle = "bba"
Output: -1
Clarification:

What should we return when needle is an empty string? This is a great question to ask during an interview.

For the purpose of this problem, we will return 0 when needle is an empty string. This is consistent to C's strstr() and Java's indexOf().

思路:快慢指针

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class Solution(object):
def strStr(self, haystack, needle):
"""
:type haystack: str
:type needle: str
:rtype: int
"""
if len(needle) == 0:
return 0
if len(haystack) == 0:
return -1
for i in range(len(haystack)):
if haystack[i] == needle[0]:
if i + len(needle) - 1 < len(haystack):
if needle == haystack[i: i + len(needle)]:
return i
else:
return -1
return -1

总结: 思路是双指针没问题,实际用 python 的时候可以用 python 的性质直接取子串

二刷:

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class Solution:
def strStr(self, haystack: str, needle: str) -> int:
if len(needle) == 0:
return 0
if len(haystack) == 0:
return -1
end = len(haystack) - len(needle) + 1
if end < 0:
return -1
for i in range(0, end):
if haystack[i:i + len(needle)] == needle:
return i
return -1

总结:注意空串的时候要返回 int 而不是 bool, needle 为空时,直接返回 0, 优化 end = len(haystack) - len(needle) + 1; if end < 0: return -1; for i in range(0, end)
高频:考点 end = lh - ln + 1; … if haystack[i:i + ln] == needle: return i。代码能优化一点点,但是大同小异。

283. Move Zeroes (Easy)

Given an array nums, write a function to move all 0’s to the end of it while maintaining the relative order of the non-zero elements.

Example:
Input: [0,1,0,3,12]
Output: [1,3,12,0,0]

Note:
You must do this in-place without making a copy of the array.
Minimize the total number of operations.

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class Solution:
def moveZeroes(self, nums: List[int]) -> None:
n = len(nums)
l, r = 0, 0
while r < n:
while nums[r] == 0 and r < n - 1:
r += 1
nums[l], nums[r] = nums[r], nums[l]
l += 1
r += 1

总结:三刷出了更简洁的写法,跑测试用例的时候仍然要考虑 0 在左中右三种情况
高频:两个月没刷这题,导致思路僵化在一刷的l, r均从0,0开始需要考虑很多种情况的算法。三刷或者l, r从0,1开始代码就简洁很多(本质是一样的)。删掉一刷代码的复杂情况判断。 非要l,r从0,0开始,就不考虑各种情况,l,r永远前进
五刷:写出了也能过的代码,但是不如高频代码简洁

125. Valid Palindrome (Easy)

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Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.

Note: For the purpose of this problem, we define empty string as valid palindrome.

Example 1:

Input: "A man, a plan, a canal: Panama"
Output: true
Example 2:

Input: "race a car"
Output: false

思路:头尾双指针, 碰头了返回 True,相同继续走,不同返回 False

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class Solution(object):
def isPalindrome(self, s):
"""
:type s: str
:rtype: bool
"""
if len(s) == 0 or len(s) == 1:
return True
head, tail = 0, len(s) - 1
while head < tail:
while not s[head].isalnum() and head < tail:
head += 1
while not s[tail].isalnum() and head < tail:
tail -= 1
if s[head].lower() != s[tail].lower():
return False
else:
head += 1
tail -= 1
return True

总结:思路简单, 但是要想到的 case 很多。考虑带标点符号,连续两个位置都是标点符号,整个字符串都是标点符合这三个情况才能写对

二刷:

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class Solution:
def isPalindrome(self, s: str) -> bool:
if len(s) <= 1:
return True
head, tail = 0, len(s) - 1
while head < tail:
while not s[head].isalnum() and head < tail:
head += 1
while not s[tail].isalnum() and head < tail:
tail -= 1
if s[head].lower() == s[tail].lower():
head += 1
tail -= 1
else:
return False
return True

总结:关键是知道 .isalnum() 这个 function
高频

680. Valid Palindrome II (Easy)

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Given a non-empty string s, you may delete at most one character. Judge whether you can make it a palindrome.

Example 1:
Input: "aba"
Output: True
Example 2:
Input: "abca"
Output: True
Explanation: You could delete the character 'c'.
Note:
The string will only contain lowercase characters a-z. The maximum length of the string is 50000.

思路:目前网上看到大部分答案都以贪心算法为主,等看贪心了再刷这题。再看一眼感觉就是统计有没有 > 2 单数的题,撸之

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class Solution(object):
def validPalindrome(self, s):
"""
:type s: str
:rtype: bool
"""
if len(s) <= 2:
return True
l, r = 0, len(s) - 1
while l < r:
if s[l] != s[r]:
break
l += 1
r -= 1
if l >= r:
return True
# 要么删左边,要么删右边
return self.isPalindrome(s[:l] + s[l + 1:]) or self.isPalindrome(s[:r] + s[r + 1:])
def isPalindrome(self, s):
return s == s[::-1]

总结:没那么简单,还要考虑这些情况 1.如果有 2 个 single 均不在 mid 位置;2. 去掉 single 点后的 string 仍然不是 palindrome; 3. 1 个 single,多个位置可以删除; 然后就抓狂了。 看了答案, 真他妈的妖。双指针算法。从两头走到中间,发现第一对不一样的字符之后,要么删左边的,要么删右边的。

二刷:

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class Solution:
def validPalindrome(self, s: str) -> bool:
if len(s) <= 2:
return True
l, r = 0, len(s) - 1
while l < r:
if s[l] != s[r]:
break
l += 1
r -= 1
if not l < r:
return True
return self.isPalindrome(s[:l] + s[l + 1:]) or self.isPalindrome(s[:r] + s[r + 1:])
def isPalindrome(self, s):
return s == s[::-1]

总结:如一刷所说,是一道比较妖的题,背熟吧

1. Two Sum (Easy)

Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

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class Solution:
def twoSum(self, nums, target):
for i in range(len(nums)):
for j in range(i + 1, len(nums)):
if nums[i] + nums[j] == target:
return [i, j]

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class Solution(object):
def twoSum(self, nums, target):
remain = {}
for i, n in enumerate(nums):
if n in remain:
return [remain[n], i]
remain[target - n] = i

一刷,高频,面经:维萨。简化代码, 双指针, hashmap

167. Two Sum II - Input array is sorted (Easy)

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Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2.

Note:

Your returned answers (both index1 and index2) are not zero-based.
You may assume that each input would have exactly one solution and you may not use the same element twice.
Example:

Input: numbers = [2,7,11,15], target = 9
Output: [1,2]
Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.

思路:增加了 sorted 这个条件, 第一感觉是可以折半查找了。固定 index1,index2 用折半查找获得

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class Solution:
def twoSum(self, numbers, target):
"""
:type numbers: List[int]
:type target: int
:rtype: List[int]
"""
for index1 in range(len(numbers)):
start, end = index1 + 1, len(numbers) - 1
while start + 1 < end:
mid = start + (end - start) // 2
if numbers[index1] + numbers[mid] == target:
return [index1 + 1, mid + 1]
elif numbers[index1] + numbers[mid] < target:
start = mid + 1
else:
end = mid - 1
if numbers[index1] + numbers[start] == target:
return [index1 + 1, start + 1]
elif numbers[index1] + numbers[end] == target:
return [index1 + 1, end + 1]

总结:要细心。1.题中 answers are not zero-based 2.要测两个情况 [2, 7, 19], 9 和 [5, 25, 75] 可以测出代码的问题

二刷:

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class Solution:
def twoSum(self, numbers: List[int], target: int) -> List[int]:
l, r = 0, len(numbers) - 1
while l < r:
tsum = numbers[l] + numbers[r]
if tsum == target:
return [l + 1, r + 1]
if tsum < target:
l += 1
else:
r -= 1

总结:二分法跑分不如直接双指针,可能是测试数据导致。双指针代码也简单很多

LinC 607. Two Sum III - Data structure design (Easy)

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Design and implement a TwoSum class. It should support the following operations: add and find.

add - Add the number to an internal data structure.
find - Find if there exists any pair of numbers which sum is equal to the value.

Example
add(1); add(3); add(5);
find(4) // return true
find(7) // return false

思路:add 的时候把 sum 都存 dict 里面, 查的时候直接返回 dict 里面有没有 sum. 会超时。

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class TwoSum:
keys = {}
"""
@param: number: An integer
@return: nothing
"""
def add(self, number):
# write your code here
if number not in self.keys:
self.keys[number] = 1
else:
self.keys[number] = 2
"""
@param: value: An integer
@return: Find if there exists any pair of numbers which sum is equal to the value.
"""
def find(self, value):
# write your code here
for key in self.keys:
if value - key in self.keys:
if value - key == key:
if self.keys[key] == 2:
return True
else:
return True
return False

总结:虽然是一道容易题, 第一反应的思路会超时。 需要在 find 的时候判断能凑出答案的另一个 key 是不是已经在 keys 里了。而不是先存好 sum。 还要判断两个数相同的时候有没有存过两个数。

15. 3Sum (Medium)

Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.
Note:
The solution set must not contain duplicate triplets.

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Example:
Given array nums = [-1, 0, 1, 2, -1, -4],
A solution set is:
[
[-1, 0, 1],
[-1, -1, 2]
]

暴力,会超时:

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class Solution:
def threeSum(self, nums):
remain = {}
ans = set()
nums.sort()
for i in range(len(nums) - 2):
if i and nums[i] == nums[i - 1]:
continue
for j in range(i + 1, len(nums)):
for k in range(j + 1, len(nums)):
if len(set([i, j, k])) == len([i, j, k]):
twoSum = nums[i] + nums[j]
remain[0 - twoSum] = (i, j)
if nums[k] in remain:
a, b = remain[nums[k]]
if len(set([a, b, k])) == len([a, b, k]):
ans.add(tuple(sorted([nums[a], nums[b], nums[k]])))
return list(ans)

四刷:

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class Solution:
def threeSum(self, nums):
nums.sort()
n = len(nums)
ans = []
for i in range(n - 2):
if i and nums[i] == nums[i - 1] :
continue
l, r = i + 1, n - 1
while l < r:
tSum = nums[i] + nums[l] + nums[r]
if tSum < 0:
l += 1
elif tSum > 0:
r -= 1
else:
ans.append([nums[i], nums[l], nums[r]])
while l < r and nums[l] == nums[l + 1]:
l += 1
while l < r and nums[r] == nums[r - 1]:
r -= 1
l += 1
r -= 1
return ans

面经:维萨。

LinC 382. Triangle Count (Medium)

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Given an array of integers, how many three numbers can be found in the array, so that we can build an triangle whose three edges length is the three numbers that we find?

Example
Given array S = [3,4,6,7], return 3. They are:

[3,4,6]
[3,6,7]
[4,6,7]
Given array S = [4,4,4,4], return 4. They are:

[4(1),4(2),4(3)]
[4(1),4(2),4(4)]
[4(1),4(3),4(4)]
[4(2),4(3),4(4)]

思路: 判断能不能做三角形以后全排列

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class Solution:
"""
@param S: A list of integers
@return: An integer
"""
def triangleCount(self, S):
# write your code here
S.sort(reverse=True)
sum = 0
for index1, longest in enumerate(S):
head, tail = index1 + 1, index1 + 2
while tail < len(S) and S[head] + S[tail] > longest:
tail += 1
tail -= 1
while head < tail:
sum += tail - head
head += 1
while head < tail and S[head] + S[tail] <= longest:
tail -= 1
return sum

总结:看清题目,问的是有多少个这样的三角形, 返回数就行。 全排列效率比较低。 更优解是每次定下最长边, 寻找符合条件的另外两个边的数量。 双指针的解法是将 tail 推到最小不能组成三角形的位置, 退一步, 然后从 tail 到 head 的位置的都可以组, 因为他们相加只会比最长边更长。 然后将 head 进一步(缩短),tail 边加长到大于最长边的位置,新 tail 到 head 的位置又都可以组。

16. 3Sum Closest (Medium)

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Given an array nums of n integers and an integer target, find three integers in nums such that the sum is closest to target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

Example:

Given array nums = [-1, 2, 1, -4], and target = 1.

The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
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class Solution:
def threeSumClosest(self, nums: List[int], target: int) -> int:
ans = None
nums.sort()
for i in range(len(nums) - 2):
l, r = i + 1, len(nums) - 1
while l < r:
t = nums[i] + nums[l] + nums[r]
if t == target:
return t
elif t < target:
l += 1
else:
r -= 1
if ans == None or abs(t - target) < abs(ans - target):
ans = t
return ans

高频:将1,2刷的代码思路总结都删了,都差不多。注意这里没有mid,不是二分查找。

LinC 31. Partition Array (Medium)

Description
Given an array nums of integers and an int k, partition the array (i.e move the elements in “nums”) such that:
All elements < k are moved to the left
All elements >= k are moved to the right
Return the partitioning index, i.e the first index i nums[i] >= k.
You should do really partition in array nums instead of just counting the numbers of integers smaller than k.
If all elements in nums are smaller than k, then return nums.length

Example
If nums = [3,2,2,1] and k=2, a valid answer is 1.

Challenge
Can you partition the array in-place and in O(n)?

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class Solution:
def partitionArray(self, nums, k):
if len(nums) == 0:
return 0
l, r = 0, len(nums) - 1
while l <= r:
while l < len(nums) and nums[l] < k:
l += 1
while r >= 0 and nums[r] >= k:
r -= 1
if l > r:
break
nums[l], nums[r] = nums[r], nums[l]
return l

总结:注意:1。while 的条件是 l <= r 2.l > r 的时候需要 break

215. Kth Largest Element in an Array (Medium)

Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.
Example 1:
Input: [3,2,1,5,6,4] and k = 2
Output: 5

Example 2:
Input: [3,2,3,1,2,4,5,5,6] and k = 4
Output: 4
Note:
You may assume k is always valid, 1 ≤ k ≤ array’s length.

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class Solution(object):
def findKthLargest(self, nums, k):
return self.quickSelect(nums, 0, len(nums) - 1, k - 1)
def quickSelect(self, nums, l, r, k):
hi = l
from random import randint
rand = randint(l, r)
nums[r], nums[rand] = nums[rand], nums[r]
for i in range(l, r):
if nums[i] > nums[r]:
nums[hi], nums[i] = nums[i], nums[hi]
hi += 1
nums[hi], nums[r] = nums[r], nums[hi]
if hi > k:
return self.quickSelect(nums, l, hi - 1, k)
elif hi < k:
return self.quickSelect(nums, hi + 1, r, k)
else:
return nums[hi]

思路:quickselect算法,基于quicksort
总结:递归调用的时候函数名前要加 return 否则不会返回任何值。由于完全抛弃另一侧,时间复杂度平均由 quick sort 的 O(nlogn) 降为 O(n) 因为输入变小了, quicksort 的输入一直是 n, 最差情况 O(n^2)
二刷:删掉一刷代码。递归调用的时候记得函数名前要加 return 否则不会返回任何值。…hi = l…for i, v in enumerate(nums[l:r], l):…
面经:维萨

75. Sort Colors (Medium)

Given an array with n objects colored red, white or blue, sort them in-place so that objects of the same color are adjacent, with the colors in the order red, white and blue.
Here, we will use the integers 0, 1, and 2 to represent the color red, white, and blue respectively.

Note: You are not suppose to use the library’s sort function for this problem.

Example:
Input: [2,0,2,1,1,0]
Output: [0,0,1,1,2,2]
Follow up:

A rather straight forward solution is a two-pass algorithm using counting sort.
First, iterate the array counting number of 0’s, 1’s, and 2’s, then overwrite array with total number of 0’s, then 1’s and followed by 2’s.
Could you come up with a one-pass algorithm using only constant space?

二刷:

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class Solution:
def sortColors(self, nums: List[int]) -> None:
"""
Do not return anything, modify nums in-place instead.
"""
if len(nums) <= 1:
return
l, r, i = 0, len(nums) - 1, 0
while i <= r:
if nums[i] == 0 and i > l:
nums[l], nums[i] = nums[i], nums[l]
l += 1
elif nums[i] == 2 and i < r:
nums[i], nums[r] = nums[r], nums[i]
r -= 1
else:
i += 1

总结:in place 不数元素的话得用 l, r 和 i, 要过的话需要熟记交换的第二条件分别为 i > l 和 i < r, 其他情况 i 均前进
高频:去掉了一刷繁琐的方法。counting sort只需要count 0和1。1 pass:…while i <= r:…and i > l:…and i < r:…
面经:Celo。3个数要保持两个边界l和r,和一个worker i,交换条件要加…i > l…和…i < r,否则会过度交换导致结果有bug

18. 4Sum (Medium)

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Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note:

The solution set must not contain duplicate quadruplets.

Example:

Given array nums = [1, 0, -1, 0, -2, 2], and target = 0.

A solution set is:
[
[-1, 0, 0, 1],
[-2, -1, 1, 2],
[-2, 0, 0, 2]
]

思路:看了下三年前的答案,不是特别直观。看了九章的答案,貌似好理解一点:去重,枚举一个数,然后用 3Sum 的做法,O(N^3)

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class Solution:
def fourSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[List[int]]
"""
nums.sort()
ans = []
for i in range(0, len(nums) - 3):
if i and nums[i] == nums[i - 1]:
continue
for j in range(i + 1, len(nums) - 2):
if j != i + 1 and nums[j] == nums[j - 1]:
continue
l, r = j + 1, len(nums) - 1
while l < r:
sum = nums[i] + nums[j] + nums[l] + nums[r]
if sum == target:
ans.append([nums[i], nums[j], nums[l], nums[r]])
l += 1
r -= 1
while l < r and nums[l] == nums[l - 1]:
l += 1
while l < r and nums[r] == nums[r + 1]:
r -= 1
elif sum < target:
l += 1
else:
r -= 1
return ans

总结:有一个自己肯定想不出的条件就是第二层循环怎么跳过:if j != i + 1 and nums[j] == nums[j - 1]: continue; 非常勉强能过 AC. 看了网上和三年前的,都是用 dict 先存 2sum,然后再 loop 两遍,用 if pair[0] > j 来去重(第三个元素的 index 要大于前面两个)。

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class Solution:
def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
ans = []
def nsum(l, r, N, target, path):
if r - l + 1 < N or N < 2 or N > len(nums) or N * nums[l] > target or N * nums[r] < target:
return
if N == 2:
while l < r:
t = nums[l] + nums[r]
if t == target:
ans.append(path + [nums[l], nums[r]])
while l < r and nums[l] == nums[l + 1]:
l += 1
while l < r and nums[r] == nums[r - 1]:
r -= 1
l += 1
r -= 1
elif t < target:
l += 1
else:
r -= 1
else:
for i in range(l, r + 1):
if i == l or (i > l and nums[i] != nums[i - 1]):
nsum(i + 1, r, N - 1, target - nums[i], path + [nums[i]])
nums.sort()
nsum(0, len(nums) - 1, 4, target, [])
return ans

二刷:看 leetcode ac 的流行答案, 返回递归 nsum, 递归内终结条件为解决 2sum,,注意两处去重,1.找到 target 以后,在 l < r 条件下跳过所有后面与 l 相同的;2.进入 nsum 前,if i == 0 or (i > 0 and nums[i - 1] != nums[i])
总结:很多坑,N == 2 时要注意 while l < r 做二分法;N > 2 时 for i in range(l, r + 1); nsum(i + 1, …); 如是高频题需要练熟
高频:…def nsum(l, r, N, target, path): if r - l + 1 < N or N < 2 or…if N == 2: while l < r:… while l < r and nums[l] == nums[l + 1]:…while…l += 1; r -= 1… for i in range(l, r + 1): if (i == l) or…: nsum(i + 1, r…)…

27. Remove Element

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Given an array nums and a value val, remove all instances of that value in-place and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

The order of elements can be changed. It doesn't matter what you leave beyond the new length.

Example 1:

Given nums = [3,2,2,3], val = 3,

Your function should return length = 2, with the first two elements of nums being 2.

It doesn't matter what you leave beyond the returned length.
Example 2:

Given nums = [0,1,2,2,3,0,4,2], val = 2,

Your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4.

Note that the order of those five elements can be arbitrary.

It doesn't matter what values are set beyond the returned length.

高频

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class Solution:
def removeElement(self, nums: List[int], val: int) -> int:
ans = len(nums)
i = 0
j = ans - 1
while i <= j:
while i <= j and nums[i] != val:
i += 1
while i <= j and nums[j] == val:
j -= 1
ans -= 1
if i < j:
nums[i], nums[j] = nums[j], nums[i]
return ans

总结:背while i <= j: while i <= j and … while i <= j and …if i < j: …

11. Container With Most Water (Medium)

Given n non-negative integers a1, a2, …, an , where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.

Note: You may not slant the container and n is at least 2.
Container With Most Water example
The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.

Example:
Input: [1,8,6,2,5,4,8,3,7]
Output: 49

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class Solution:
def maxArea(self, height: List[int]) -> int:
ans, l, r = 0, 0, len(height) - 1
while l < r:
ans = max(ans, (r - l) * min(height[l], height[r]))
if height[l] < height[r]:
l += 1
else:
r -= 1
return ans

高频, 二刷:1.短的不能贡献更大的面积,可以移除2.其他的都需要更高的才有可能变成更大面积

345. Reverse Vowels of a String (Easy)

Write a function that takes a string as input and reverse only the vowels of a string.

Example 1:
Input: “hello”
Output: “holle”

Example 2:
Input: “leetcode”
Output: “leotcede”
Note:

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class Solution:
def reverseVowels(self, s: str) -> str:
stack = []
for c in s:
if c in "aeiouAEIOU":
stack.append(c)
for i, c in enumerate(s):
if c in "aeiouAEIOU":
s = s[:i] + stack.pop() + s[i + 1:]
return s

inplace:

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class Solution:
def reverseVowels(self, s: str) -> str:
l, r = 0, len(s) - 1
arr = list(s)
while l < r:
while arr[l] not in "aeiouAEIOU" and l < len(s) - 1:
l += 1
while arr[r] not in "aeiouAEIOU" and r > 0:
r -= 1
if l < r:
arr[l], arr[r] = arr[r], arr[l]
l += 1
r -= 1
return "".join(arr)

面经:DJI。

BFS 广度优先搜索

695. Max Area of Island (Medium)

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Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

Find the maximum area of an island in the given 2D array. (If there is no island, the maximum area is 0.)

Example 1:
[[0,0,1,0,0,0,0,1,0,0,0,0,0],
[0,0,0,0,0,0,0,1,1,1,0,0,0],
[0,1,1,0,1,0,0,0,0,0,0,0,0],
[0,1,0,0,1,1,0,0,1,0,1,0,0],
[0,1,0,0,1,1,0,0,1,1,1,0,0],
[0,0,0,0,0,0,0,0,0,0,1,0,0],
[0,0,0,0,0,0,0,1,1,1,0,0,0],
[0,0,0,0,0,0,0,1,1,0,0,0,0]]
Given the above grid, return 6. Note the answer is not 11, because the island must be connected 4-directionally.
Example 2:
[[0,0,0,0,0,0,0,0]]
Given the above grid, return 0.
Note: The length of each dimension in the given grid does not exceed 50.

思路:没啥太多好说的,BFS 暴力

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class Solution:
def maxAreaOfIsland(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
if len(grid) == 0:
return 0
ans = 0
for row in range(len(grid)):
for col in range(len(grid[0])):
if grid[row][col] == 1:
ans = max(self.bfs(grid, row, col), ans)
return ans
def bfs(self, grid, row, col):
import collections
q = collections.deque()
q.append((row, col))
grid[row][col] = 0
size = 0
while q:
row, col = q.popleft()
size += 1
if self.isValid(grid, row - 1, col) and grid[row - 1][col] == 1:
q.append((row - 1, col))
grid[row - 1][col] = 0
if self.isValid(grid, row + 1, col) and grid[row + 1][col] == 1:
q.append((row + 1, col))
grid[row + 1][col] = 0
if self.isValid(grid, row, col - 1) and grid[row][col - 1] == 1:
q.append((row, col - 1))
grid[row][col - 1] = 0
if self.isValid(grid, row, col + 1) and grid[row][col + 1] == 1:
q.append((row, col + 1))
grid[row][col + 1] = 0
return size
def isValid(self, grid, row, col):
return row >= 0 and row < len(grid) and col >= 0 and col < len(grid[0])

总结:在上下左右走的时候注意入 q 以后立刻将该点标为 0, 以防同一个点入两次。
二刷:

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class Solution:
def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
if len(grid) == 0 or len(grid[0]) == 0:
return 0
ans = 0
for r in range(len(grid)):
for c in range(len(grid[0])):
if grid[r][c] == 1:
s = 0
q = collections.deque()
q.append((r, c))
grid[r][c] = 0
while len(q) > 0:
(rq, cq) = q.popleft()
s += 1
if rq > 0 and grid[rq - 1][cq] == 1:
q.append((rq - 1, cq))
grid[rq - 1][cq] = 0
if rq < len(grid) - 1 and grid[rq + 1][cq] == 1:
q.append((rq + 1, cq))
grid[rq + 1][cq] = 0
if cq > 0 and grid[rq][cq - 1] == 1:
q.append((rq, cq - 1))
grid[rq][cq - 1] = 0
if cq < len(grid[0]) - 1 and grid[rq][cq + 1] == 1:
q.append((rq, cq + 1))
grid[rq][cq + 1] = 0
if s > ans:
ans = s
return ans

总结:一次写完, 没有像一刷那样拆成三个函数。 各有优劣吧。注意清零的位置要在放 queue 之后立刻清零,以防同一个点如两次, lol 二刷踩同样的坑 :’(

133. Clone Graph (Medium)

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Given a reference of a node in a connected undirected graph, return a deep copy (clone) of the graph. Each node in the graph contains a val (int) and a list (List[Node]) of its neighbors.

Example:
<img src="https://assets.leetcode.com/uploads/2019/02/19/113_sample.png"
alt="graph example"/>
Input:
{"$id":"1","neighbors":[{"$id":"2","neighbors":[{"$ref":"1"},{"$id":"3","neighbors":[{"$ref":"2"},{"$id":"4","neighbors":[{"$ref":"3"},{"$ref":"1"}],"val":4}],"val":3}],"val":2},{"$ref":"4"}],"val":1}

Explanation:
Node 1's value is 1, and it has two neighbors: Node 2 and 4.
Node 2's value is 2, and it has two neighbors: Node 1 and 3.
Node 3's value is 3, and it has two neighbors: Node 2 and 4.
Node 4's value is 4, and it has two neighbors: Node 1 and 3.

Note:
The number of nodes will be between 1 and 100.
The undirected graph is a simple graph, which means no repeated edges and no self-loops in the graph.
Since the graph is undirected, if node p has node q as neighbor, then node q must have node p as neighbor too.
You must return the copy of the given node as a reference to the cloned graph.

思路:BFS, 用一个 dict 存当前节点的邻居,如果没见过就加 dict 存 queue,queue 出来建 node,放 neighbor;概念上比较好懂,写码可能有坑

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# Definition for a undirected graph node
# class UndirectedGraphNode:
# def __init__(self, x):
# self.label = x
# self.neighbors = []

class Solution:
# @param node, a undirected graph node
# @return a undirected graph node
def cloneGraph(self, node):
if node == None:
return None
dict = {}
_cloneNode = UndirectedGraphNode(node.label)
dict[node] = _cloneNode
q = [node]
while q:
new_q = []
for _node in q:
for neighbor in _node.neighbors:
if neighbor not in dict:
_cloneNode = UndirectedGraphNode(neighbor.label)
dict[neighbor] = _cloneNode
new_q.append(neighbor)
dict[_node].neighbors.append(dict[neighbor])
q = new_q
return dict[node]

总结:思路用 dict 来存当前节点的邻居是错的,需要用 dict 存当前节点和克隆节点的映射关系。因为反正映射关系在,加邻居可以后加. 邻居是不能直接 copy 或者 = 的, 因为邻居的类型也是节点, 需要创造以后加进去。测一下,然后 debug 细一点, 要测出

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dict[_node].neighbors.append(dict[neighbor])

二刷:

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"""
# Definition for a Node.
class Node:
def __init__(self, val, neighbors):
self.val = val
self.neighbors = neighbors
"""
class Solution:
def cloneGraph(self, node: 'Node') -> 'Node':
d = {}
n = Node(node.val, [])
if node.neighbors == None or len(node.neighbors) == 0:
return n
d[node] = n
q = collections.deque()
q.append(node)
while q:
nq = q.popleft()
if nq not in d:
n = Node(nq.val, [])
d[nq] = n
for nn in nq.neighbors:
if nn not in d:
q.append(nn)
n = Node(nn.val, [])
d[nn] = n
d[nq].neighbors.append(d[nn])
return d[node]

总结:二刷第一遍没有想到 nn 在不在 d 里面,都要加入到克隆出来的节点的 neighbors 中去。此次击败了 100% 的内存使用,如需提高速度,可以增加一个 visited = set() 如果已经访问过就 return,可以用空间换时间

127. Word Ladder (Medium)

Given two words (beginWord and endWord), and a dictionary’s word list, find the length of shortest transformation sequence from beginWord to endWord, such that:
Only one letter can be changed at a time.
Each transformed word must exist in the word list. Note that beginWord is not a transformed word.
Note:
Return 0 if there is no such transformation sequence.
All words have the same length.
All words contain only lowercase alphabetic characters.
You may assume no duplicates in the word list.
You may assume beginWord and endWord are non-empty and are not the same.

Example 1:
Input:
beginWord = “hit”,
endWord = “cog”,
wordList = [“hot”,”dot”,”dog”,”lot”,”log”,”cog”]
Output: 5
Explanation: As one shortest transformation is “hit” -> “hot” -> “dot” -> “dog” -> “cog”,
return its length 5.

Example 2:
Input:
beginWord = “hit”
endWord = “cog”
wordList = [“hot”,”dot”,”dog”,”lot”,”log”]
Output: 0
Explanation: The endWord “cog” is not in wordList, therefore no possible transformation.

可以输出此path:

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class Solution:
def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
if endWord not in wordList or len(beginWord) != len(endWord):
return 0
q = [[beginWord]]
wordList = set(wordList)
visited = set()
while q:
curPath = q.pop(0)
curWord = curPath[-1]
if curWord == endWord:
return len(curPath)
for i in range(len(curWord)):
for c in [chr(x) for x in range(ord("a"), ord("z") + 1)]:
newW = curWord[:i] + c + curWord[i + 1:]
if newW in wordList and newW not in visited:
visited.add(newW)
q.append(curPath + [newW])
return 0

精简无需输出path:

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class Solution:
def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
if endWord not in wordList or len(beginWord) != len(endWord):
return 0
q = [(beginWord, 1)]
wordList = set(wordList)
visited = set()
while q:
word, length = q.pop(0)
if word == endWord:
return length
for i in range(len(word)):
for c in [chr(i) for i in range(ord("a"), ord("z"))]:
newW = word[:i] + c + word[i + 1:]
if newW in wordList and newW not in visited:
visited.add(newW)
q.append((newW, length + 1))
return 0

面试:DJI
三刷:注意简版需要visited

200. Number of Islands (Medium)

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Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1:
Input:
11110
11010
11000
00000
Output: 1

Example 2:
Input:
11000
11000
00100
00011
Output: 3

思路:遍历矩阵,碰到 1 就上下左右 BFS,碰到 0 跳过。BFS 访问过的标 0

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class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
ans = 0
for r in range(len(grid)):
for c in range(len(grid[0])):
if grid[r][c] == "1":
ans += 1
grid[r][c] = "0"
self.bfs(grid, r, c)
return ans
def bfs(self, matrix, r, c):
q = [(r, c)]
while q:
r, c = q.pop(0)
for dr, dc in [(1, 0), (-1, 0), (0, 1), (0, -1)]:
newR = r + dr
newC = c + dc
if 0 <= newR <= len(matrix) - 1 and 0 <= newC <= len(matrix[0]) - 1 and matrix[newR][newC] == "1":
matrix[newR][newC] = "0"
q.append((newR, newC))

面经,三刷:Amazon。对于 leetcode ac 比较重要的细节是,gird[][] = ‘0’ 这句话要在 if 里面,否则逻辑 OK 但是会 TLE

LinC 611. Knight Shortest Path (Medium)

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Given a knight in a chessboard (a binary matrix with 0 as empty and 1 as barrier) with a source position, find the shortest path to a destination position, return the length of the route.
Return -1 if knight can not reached.

source and destination must be empty.
Knight can not enter the barrier.

Clarification
If the knight is at (x, y), he can get to the following positions in one step:

(x + 1, y + 2)
(x + 1, y - 2)
(x - 1, y + 2)
(x - 1, y - 2)
(x + 2, y + 1)
(x + 2, y - 1)
(x - 2, y + 1)
(x - 2, y - 1)
Example
[[0,0,0],
[0,0,0],
[0,0,0]]
source = [2, 0] destination = [2, 2] return 2

[[0,1,0],
[0,0,0],
[0,0,0]]
source = [2, 0] destination = [2, 2] return 6

[[0,1,0],
[0,0,1],
[0,0,0]]
source = [2, 0] destination = [2, 2] return -1

思路:没什么思路, 看了下答案,就是 BFS 硬来,需要检查走了某个方向以后是不是还是在棋盘内

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"""
Definition for a point.
class Point:
def __init__(self, a=0, b=0):
self.x = a
self.y = b
"""

class Solution:
"""
@param grid: a chessboard included 0 (false) and 1 (true)
@param source: a point
@param destination: a point
@return: the shortest path
"""
def shortestPath(self, grid, source, destination):
# write your code here
if len(grid) == 0 or (len(grid[0]) == 1 and grid[0][0] == 1):
return -1
ans = 0
dx = [1, 1, -1, -1, 2, 2, -2, -2]
dy = [2, -2, 2, -2, 1, -1, 1, -1]
q = collections.deque([source])
grid[source.x][source.y] = 1
while q:
qlen = len(q)
next_q = collections.deque()
for i in range(qlen):
pt = q.popleft()
if pt.x == destination.x and pt.y == destination.y:
return ans
for move in range(len(dx)):
nextPt = Point(pt.x + dx[move], pt.y + dy[move])
if (self.isInbound(grid, nextPt) and grid[nextPt.x][nextPt.y] == 0):
next_q.append(nextPt)
grid[nextPt.x][nextPt.y] = 1
ans += 1
q = next_q
return -1
def isInbound(self, grid, pt):
return pt.x >= 0 and pt.x < len(grid) and pt.y >= 0 and pt.y < len(grid[0])

总结:注意 isInbound 要查的是 >=0 和 < len(), 其他的问题可以通过跑一个测试数据发现

785. Is Graph Bipartite? (Medium)

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Given an undirected graph, return true if and only if it is bipartite.

Recall that a graph is bipartite if we can split it's set of nodes into two independent subsets A and B such that every edge in the graph has one node in A and another node in B.

The graph is given in the following form: graph[i] is a list of indexes j for which the edge between nodes i and j exists. Each node is an integer between 0 and graph.length - 1. There are no self edges or parallel edges: graph[i] does not contain i, and it doesn't contain any element twice.

Example 1:
Input: [[1,3], [0,2], [1,3], [0,2]]
Output: true
Explanation:
The graph looks like this:
0----1
| |
| |
3----2
We can divide the vertices into two groups: {0, 2} and {1, 3}.
Example 2:
Input: [[1,2,3], [0,2], [0,1,3], [0,2]]
Output: false
Explanation:
The graph looks like this:
0----1
| \ |
| \ |
3----2
We cannot find a way to divide the set of nodes into two independent subsets.


Note:

graph will have length in range [1, 100].
graph[i] will contain integers in range [0, graph.length - 1].
graph[i] will not contain i or duplicate values.
The graph is undirected: if any element j is in graph[i], then i will be in graph[j].

思路:用染色的方法,可以用 DFS, BFS 给所有 node 染上两种色中的一种。1.未上色,既上色,给相邻节点上相反色 2.已上色,查是否和目前要上的色相同
DFS:

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class Solution:
def isBipartite(self, graph):
"""
:type graph: List[List[int]]
:rtype: bool
"""
seen = {}
def dfs(n, color):
if n in seen:
return color == seen[n]
seen[n] = color
for v in graph[n]:
if not dfs(v, -color):
return False
return True
for i in range(len(graph)):
if i not in seen and dfs(i, 1) == False:
return False
return True

BFS:

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class Solution:
def isBipartite(self, graph):
seen = {}
for i in range(len(graph)):
if i not in seen:
s = [(i, 1)]
seen[i] = 1
while s:
n, color = s.pop()
for v in graph[n]:
if v in seen:
if color == seen[v]:
return False
else:
seen[v] = -color
s.append((v, -color))
return True

二刷:DFS: …return color == seen[n]…if not dfs(v, -color): return False…return True… BFS: …seen[i] = 1; while…

LinC 178. Graph Valid Tree (Medium)

Leetcode 261. Graph Valid Tree 带锁

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Given n nodes labeled from 0 to n - 1 and a list of undirected edges (each edge is a pair of nodes), write a function to check whether these edges make up a valid tree.

You can assume that no duplicate edges will appear in edges. Since all edges are undirected, [0, 1] is the same as [1, 0] and thus will not appear together in edges.

Example
Given n = 5 and edges = [[0, 1], [0, 2], [0, 3], [1, 4]], return true.

Given n = 5 and edges = [[0, 1], [1, 2], [2, 3], [1, 3], [1, 4]], return false.

思路:树的两个条件是不能有环,不能有孤儿节点。怎么实现想不太出来。看了答案,用 defaultdict(list) 放节点之间的关系, 有没有环其实不用管,因为只要确保边的数量 == n - 1, 并且 BFS 走过一遍之后访问过了所有的点,就确定没有环了。

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class Solution:
def validTree(self, n, edges):
if len(edges) == 0 and n == 1:
return True
if len(edges) != n - 1:
return False
mapping = collections.defaultdict(list)
for edge in edges:
mapping[edge[0]].append(edge[1])
mapping[edge[1]].append(edge[0])
visited = set()
q = [0]
while q:
node = q.pop()
visited.add(node)
for neighbor in mapping[node]:
if neighbor not in visited:
q.append(neighbor)
visited.add(neighbor)
return len(visited) == n

总结:相当值得做的一道 BFS 题。注意 已经访问过的节点不要入 q,不然无向图的边会导致死循环

130. Surrounded Regions (Medium)

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Given a 2D board containing 'X' and 'O' (the letter O), capture all regions surrounded by 'X'.

A region is captured by flipping all 'O's into 'X's in that surrounded region.

Example:

X X X X
X O O X
X X O X
X O X X
After running your function, the board should be:

X X X X
X X X X
X X X X
X O X X
Explanation:

Surrounded regions shouldn’t be on the border, which means that any 'O' on the border of the board are not flipped to 'X'. Any 'O' that is not on the border and it is not connected to an 'O' on the border will be flipped to 'X'. Two cells are connected if they are adjacent cells connected horizontally or vertically.
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class Solution:
def solve(self, board: List[List[str]]) -> None:
"""
Do not return anything, modify board in-place instead.
"""
if not board:
return
m, n = len(board), len(board[0])
rule = lambda ij: 0 <= ij[0] < m and 0 <= ij[1] < n and board[ij[0]][ij[1]] == 'O'
q = list(filter(rule, [ij for k in range(max(m, n)) for ij in [(0, k), (k, 0), (m - 1, k), (k, n - 1)]]))
while q:
(i, j) = q.pop()
board[i][j] = 'S'
q += list(filter(rule, [(i, j - 1), (i, j + 1), (i - 1, j), (i + 1, j)]))
for i in range(m):
for j in range(n):
if board[i][j] == 'S':
board[i][j] = 'O'
else:
board[i][j] = 'X'

高频:需要改为q.pop(0)才是BFS,否则是DFS,但是代码风格放在BFS比较合适

675. Cut Off Trees for Golf Event (Hard)

You are asked to cut off trees in a forest for a golf event. The forest is represented as a non-negative 2D map, in this map:
0 represents the obstacle can’t be reached.
1 represents the ground can be walked through.
The place with number bigger than 1 represents a tree can be walked through, and this positive number represents the tree’s height.

You are asked to cut off all the trees in this forest in the order of tree’s height - always cut off the tree with lowest height first. And after cutting, the original place has the tree will become a grass (value 1).
You will start from the point (0, 0) and you should output the minimum steps you need to walk to cut off all the trees. If you can’t cut off all the trees, output -1 in that situation.
You are guaranteed that no two trees have the same height and there is at least one tree needs to be cut off.

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Example 1:
Input:
[
[1,2,3],
[0,0,4],
[7,6,5]
]
Output: 6

Example 2:
Input:
[
[1,2,3],
[0,0,0],
[7,6,5]
]
Output: -1

Example 3:
Input:
[
[2,3,4],
[0,0,5],
[8,7,6]
]
Output: 6
Explanation: You started from the point (0,0) and you can cut off the tree in (0,0) directly without walking.

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class Solution:
def cutOffTree(self, forest):
m, n = len(forest), len(forest[0])
trees = [[forest[r][c], r, c] for r in range(m) for c in range(n) if forest[r][c] > 1]
trees.sort(key = lambda x: x[0])
ans = 0
nextR, nextC = 0, 0
for h, r, c in trees:
step = self.bfs(forest, nextR, nextC, r, c, m, n)
if step == -1:
return -1
else:
forest[r][c] = 1
nextR, nextC = r, c
ans += step
return ans
def bfs(self, forest, startR, startC, endR, endC, m, n):
step = 0
q = [(startR, startC)]
dirs = [(-1, 0), (1, 0), (0, -1), (0, 1)]
visited = set()
while q:
nq = []
for _ in range(len(q)):
r, c = q.pop()
if r == endR and c == endC:
return step
visited.add((r, c))
for dr, dc in dirs:
if 0 <= r + dr <= m - 1 and 0 <= c + dc <= n - 1 and forest[r + dr][c + dc] != 0 and (r + dr, c + dc) not in visited:
nq.append((r + dr, c + dc))
step += 1
q = nq
return -1

面经:Amazon。比较不偏门的算法,可惜会TLE

310. Minimum Height Trees (Medium)

For an undirected graph with tree characteristics, we can choose any node as the root. The result graph is then a rooted tree. Among all possible rooted trees, those with minimum height are called minimum height trees (MHTs). Given such a graph, write a function to find all the MHTs and return a list of their root labels.
Format
The graph contains n nodes which are labeled from 0 to n - 1. You will be given the number n and a list of undirected edges (each edge is a pair of labels).
You can assume that no duplicate edges will appear in edges. Since all edges are undirected, [0, 1] is the same as [1, 0] and thus will not appear together in edges.

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Example 1 :
Input: n = 4, edges = [[1, 0], [1, 2], [1, 3]]
0
|
1
/ \
2 3

Output: [1]

Example 2 :
Input: n = 6, edges = [[0, 3], [1, 3], [2, 3], [4, 3], [5, 4]]
0 1 2
\ | /
3
|
4
|
5

Output: [3, 4]

Note:
According to the definition of tree on Wikipedia: “a tree is an undirected graph in which any two vertices are connected by exactly one path. In other words, any connected graph without simple cycles is a tree.”
The height of a rooted tree is the number of edges on the longest downward path between the root and a leaf.

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class Solution:
def findMinHeightTrees(self, n: int, edges: List[List[int]]) -> List[int]:
if not edges:
return [0]
from collections import defaultdict
adj = defaultdict(list)
for u, v in edges:
adj[u].append(v)
adj[v].append(u)
leaves = [k for k, v in adj.items() if len(v) == 1]
while n > 2:
n -= len(leaves)
newLeaves = []
for u in leaves:
v = adj[u].pop()
adj[v].remove(u)
if len(adj[v]) == 1:
newLeaves.append(v)
leaves = newLeaves
return leaves

一刷:Facebook tag,leetcode 讨论区解法

Topological sorting 拓扑排序

LinC 127. Topological Sorting (Medium)

Given an directed graph, a topological order of the graph nodes is defined as follow:

For each directed edge A -> B in graph, A must before B in the order list.
The first node in the order can be any node in the graph with no nodes direct to it.
Find any topological order for the given graph.
You can assume that there is at least one topological order in the graph.
Clarification
Learn more about representation of graphs

Example:
For graph as follow:
graph example
The topological order can be:
[0, 1, 2, 3, 4, 5]
[0, 2, 3, 1, 5, 4]
思路:拓扑排序,算法貌似是:1.统计每个点的入度;2.将入度为 0 的点入 queue;3.从队列中 pop 点,去掉所有指向别的点的边: 相应点入度 -1;4.新入度为 0 的点入 queue

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"""
Definition for a Directed graph node
class DirectedGraphNode:
def __init__(self, x):
self.label = x
self.neighbors = []
"""
class Solution:
"""
@param: graph: A list of Directed graph node
@return: Any topological order for the given graph.
"""
def topSort(self, graph):
# write your code here
if len(graph) == 0:
return []
ans = []
inBound = {}
for node in graph:
if node not in inBound:
inBound[node] = 0
for neighbor in node.neighbors:
if neighbor not in inBound:
inBound[neighbor] = 0
inBound[neighbor] += 1
q = collections.deque()
for node in inBound:
if inBound[node] == 0:
q.append(node)
while q:
zNode = q.popleft()
ans.append(zNode)
for node in zNode.neighbors:
inBound[node] -= 1
if inBound[node] == 0:
q.append(node)
return ans

总结:顺利。但是题目没有说清楚 return 的是一个拓扑排序好的 node 的 list

207. Course Schedule (Medium)

There are a total of n courses you have to take, labeled from 0 to n-1.
Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]
Given the total number of courses and a list of prerequisite pairs, is it possible for you to finish all courses?

Example 1:
Input: 2, [[1,0]]
Output: true
Explanation: There are a total of 2 courses to take.
To take course 1 you should have finished course 0. So it is possible.

Example 2:
Input: 2, [[1,0],[0,1]]
Output: false
Explanation: There are a total of 2 courses to take.
To take course 1 you should have finished course 0, and to take course 0 you should
also have finished course 1. So it is impossible.

Note:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
You may assume that there are no duplicate edges in the input prerequisites.

二刷:

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class Solution:
def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
inDegree = [0] * numCourses
graph = collections.defaultdict(list)
for e in prerequisites:
graph[e[1]].append(e[0])
inDegree[e[0]] += 1
q = []
for i, d in enumerate(inDegree):
if d == 0:
q.append(i)
while q:
v1 = q.pop(0)
for v2 in graph[v1]:
inDegree[v2] -= 1
if inDegree[v2] == 0:
q.append(v2)
for d in inDegree:
if d != 0:
return False
return True

总结:较值得二刷的题,发现了for i, v in enumerate(list)和for k, v in dict.items()这两种用法混淆的薄弱环节。 还有对入度和建的graph概念没有完全理解
面经:Cruise。…collecitons.default dict()…统计入度需要…inDegree[e[0]] += 1…被指向的vertex入度加一

210. Course Schedule II (Medium)

There are a total of n courses you have to take, labeled from 0 to n-1.
Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]
Given the total number of courses and a list of prerequisite pairs, return the ordering of courses you should take to finish all courses.
There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.

Example 1:
Input: 2, [[1,0]]
Output: [0,1]
Explanation: There are a total of 2 courses to take. To take course 1 you should have finished
course 0. So the correct course order is [0,1] .

Example 2:
Input: 4, [[1,0],[2,0],[3,1],[3,2]]
Output: [0,1,2,3] or [0,2,1,3]
Explanation: There are a total of 4 courses to take. To take course 3 you should have finished both
courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0.
So one correct course order is [0,1,2,3]. Another correct ordering is [0,2,1,3] .
Note:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
You may assume that there are no duplicate edges in the input prerequisites.

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class Solution:
def findOrder(self, numCourses: int, prerequisites: List[List[int]]) -> List[int]:
graph = collections.defaultdict(list)
indegree = [0] * numCourses
for e in prerequisites:
graph[e[1]].append(e[0])
indegree[e[0]] += 1
q = []
ans = []
for i, v in enumerate(indegree):
if v == 0:
q.append(i)
ans.append(i)
while q:
v1 = q.pop(0)
for v2 in graph[v1]:
indegree[v2] -= 1
if indegree[v2] == 0:
q.append(v2)
ans.append(v2)
return ans if len(ans) == numCourses else []

面经:Cruise。

DFS 深度优先搜索

Binary Tree DFS 二叉树与树上的深度优先搜索

257. Binary Tree Paths (Easy)

Given a binary tree, return all root-to-leaf paths.
Note: A leaf is a node with no children.

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Example:
Input:
1
/ \
2 3
\
5
Output: ["1->2->5", "1->3"]

Explanation: All root-to-leaf paths are: 1->2->5, 1->3

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class Solution:
def binaryTreePaths(self, root: TreeNode) -> List[str]:
if not root:
return []
ans = []
def dfs(root, cur):
if not root:
return
cur.append(str(root.val))
if not root.left and not root.right:
ans.append("->".join(cur))
dfs(root.left, cur)
dfs(root.right, cur)
del cur[-1]
dfs(root, [])
return ans

五刷:注意剪枝,因为cur是一个reference,递归过程中cur会变长,当前层完成返回上一级函数调用时需要将当前层加上的多余的枝减去

113. Path Sum II (Medium)

Given a binary tree and a sum, find all root-to-leaf paths where each path’s sum equals the given sum.
Note: A leaf is a node with no children.

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Example:
Given the below binary tree and sum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ / \
7 2 5 1
Return:
[
[5,4,11,2],
[5,8,4,5]
]

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class Solution:
def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]:
ans = []
target = sum
def dfs(root, path):
global sum
nonlocal target
if not root:
return
path.append(root.val)
if not root.left and not root.right:
if sum(path) == target:
ans.append(path[:])
dfs(root.left, path)
dfs(root.right, path)
del path[-1]
dfs(root, [])
return ans

四刷:高频,面经:Quora, 大疆。TODO 用非递归再刷

437. Path Sum III (Medium)

You are given a binary tree in which each node contains an integer value.
Find the number of paths that sum to a given value.
The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).
The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.

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Example:
root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8

10
/ \
5 -3
/ \ \
3 2 11
/ \ \
3 -2 1

Return 3. The paths that sum to 8 are:
1. 5 -> 3
2. 5 -> 2 -> 1
3. -3 -> 11

O(n^2)

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class Solution:
def pathSum(self, root: TreeNode, sum: int) -> int:
if not root:
return 0
ans = 0
def helper(root):
if not root:
return
dfs(root, 0)
helper(root.left)
helper(root.right)
def dfs(root, cur):
nonlocal sum, ans
if not root:
return
cur += root.val
if cur == sum:
ans += 1
dfs(root.left, cur)
dfs(root.right, cur)
cur -= root.val
helper(root)
return ans

DFS + Memo:

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class Solution:
def pathSum(self, root: TreeNode, sum: int) -> int:
ans = 0
memo = {0: 1}
def dfs(root, cur):
nonlocal ans, sum
if not root:
return
cur += root.val
ans += memo.get(cur - sum, 0)
memo[cur] = memo.get(cur, 0) + 1
dfs(root.left, cur)
dfs(root.right, cur)
memo[cur] -= 1
dfs(root, 0)
return ans

二刷:面经:Quora

Combination based DFS - 基于组合的深度优先搜索

78. Subsets (Medium)

Given a set of distinct integers, nums, return all possible subsets (the power set).
Note: The solution set must not contain duplicate subsets.

Example:
Input: nums = [1,2,3]
Output:
[
[3],
[1],
[2],
[1,2,3],
[1,3],
[2,3],
[1,2],
[]
]

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class Solution:
def subsets(self, nums):
if not nums:
return []
ans = []
def dfs(cur, idx):
ans.append(cur[:])
for i in range(idx, len(nums)):
cur.append(nums[i])
dfs(cur, i + 1)
del cur[-1]
dfs([], 0)
return ans

五刷:高频。O(2^n)

39. Combination Sum (Medium)

Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.
The same repeated number may be chosen from candidates unlimited number of times.
Note:
All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations.

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Example 1:
Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
[7],
[2,2,3]
]

Example 2:
Input: candidates = [2,3,5], target = 8,
A solution set is:
[
[2,2,2,2],
[2,3,3],
[3,5]
]

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class Solution:
def combinationSum(self, candidates, target):
ans = []
candidates.sort()
def dfs(cur, idx):
nonlocal target
if sum(cur) > target:
return
if sum(cur) == target:
ans.append(cur[:])
for i in range(idx, len(candidates)):
cur.append(candidates[i])
dfs(cur, i)
del cur[-1]
dfs([], 0)
return ans]

四刷:高频, 面经, Quora, Amazon。需要排序的原因是为了避免结果里[2,3,2], [3,2,2]这类情况的发生,结果里只能取当前或比当前大的数。时间复杂度为O(n^k),n是candidates的数量,k是target / min(candidates)或者用target来近似(假设min是1的话),这个上限比网上一些O(n!)的说法更低

40. Combination Sum II (Medium)

Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.
Each number in candidates may only be used once in the combination.
Note:
All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations.

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Example 1:
Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]

Example 2:
Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
[1,2,2],
[5]
]

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class Solution:
def combinationSum2(self, candidates, target):
candidates.sort()
ans = []
def dfs(cur, idx):
nonlocal target
if sum(cur) > target:
return
if sum(cur) == target:
ans.append(cur[:])
return
for i in range(idx, len(candidates)):
if i > idx and candidates[i] == candidates[i - 1]:
continue
cur.append(candidates[i])
dfs(cur, i + 1)
del cur[-1]
dfs([], 0)
return ans

四刷:高频, 面经, Amazon。O(2^n) 注意:…if i > idx and nums[i] == nums[i - 1]: continue…

216. Combination Sum III (Medium)

Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.
Note:
All numbers will be positive integers.
The solution set must not contain duplicate combinations.

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Example 1:
Input: k = 3, n = 7
Output: [[1,2,4]]

Example 2:
Input: k = 3, n = 9
Output: [[1,2,6], [1,3,5], [2,3,4]]

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class Solution:
def combinationSum3(self, k, n):
ans = []
nums = [i for i in range(1, 10)]
def dfs(cur, idx):
nonlocal k, n
if len(cur) > k:
return
if len(cur) == k and sum(cur) == n:
ans.append(cur[:])
return
for i in range(idx, len(nums)):
cur.append(nums[i])
dfs(cur, i + 1)
del cur[-1]
dfs([], 0)
return ans

二刷:面经

77. Combinations (Medium)

Given two integers n and k, return all possible combinations of k numbers out of 1 … n.

Example:
Input: n = 4, k = 2
Output:
[
[2,4],
[3,4],
[2,3],
[1,2],
[1,3],
[1,4],
]

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class Solution:
def combine(self, n: int, k: int) -> List[List[int]]:
ans = []
def dfs(cur, idx):
if len(cur) == k:
ans.append(cur[:])
for i in range(idx, n + 1):
cur.append(i)
dfs(cur, i + 1)
del cur[-1]
dfs([], 1)
return ans

二刷:面经:Amazon

131. Palindrome Partitioning (Medium)

Given a string s, partition s such that every substring of the partition is a palindrome.
Return all possible palindrome partitioning of s.

Example:
Input: “aab”
Output:
[
[“aa”,”b”],
[“a”,”a”,”b”]
]

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class Solution:
def partition(self, s: str) -> List[List[str]]:
ans = []
def dfs(cur, idx):
if idx == len(s):
ans.append(cur[:])
return
for i in range(idx, len(s)):
w = s[idx: i + 1]
if w == w[::-1]:
cur.append(w)
dfs(cur, i + 1)
del cur[-1]
dfs([], 0)
return ans

四刷:高频,O(n*(2^n))

93. Restore IP Addresses (Medium)

Given a string containing only digits, restore it by returning all possible valid IP address combinations.

Example:
Input: “25525511135”
Output: [“255.255.11.135”, “255.255.111.35”]

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class Solution:
def restoreIpAddresses(self, s: str) -> List[str]:
if len(s) > 12:
return []
ans = []
def dfs(cur, idx):
if idx == len(s) and len(cur) == 4:
ans.append(".".join(cur))
return
for i in range(idx, len(s)):
num = s[idx: i + 1]
if i - idx < 3 and int(num) <= 255 and len(num) == len(str(int(num))):
cur.append(num)
dfs(cur, i + 1)
del cur[-1]
dfs([], 0)
return ans

五刷:高频, 注意查cur的元素数量,还有num里有前缀为0但是num不是0的情况,时间复杂度是一件有趣的事情,正常情况下分割字符串是O(2^n)复杂度,但是ip地址是有限的,因此这个题有个常数的时间复杂度上限

LinC 680. Split String (Easy)

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Give a string, you can choose to split the string after one character or two adjacent characters, and make the string to be composed of only one character or two characters. Output all possible results.

Example
Given the string "123"
return [["1","2","3"],["12","3"],["1","23"]]
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class Solution:
def splitString(self, s):
if len(s) == 0:
return [[]]
ans = []
def dfs(cur, idx):
if idx == len(s):
ans.append(cur[:])
return
for i in range(1, 3):
if idx + i > len(s):
break
cur.append(s[idx:idx + i])
dfs(cur, idx + i)
del cur[-1]
dfs([], 0)
return ans

二刷:空输入的输出和leetcode有区别

90. Subsets II (Medium)

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Given a collection of integers that might contain duplicates, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

Example:

Input: [1,2,2]
Output:
[
[2],
[1],
[1,2,2],
[2,2],
[1,2],
[]
]
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class Solution:
def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
ans = []
def dfs(cur, idx):
ans.append(cur[:])
for i in range(idx, len(nums)):
if i > idx and nums[i] == nums[i - 1]:
continue
cur.append(nums[i])
dfs(cur, i + 1)
cur.pop()
nums.sort()
dfs([], 0)
return ans

四刷:高频

140. Word Break II (Hard)

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences.

Note:

The same word in the dictionary may be reused multiple times in the segmentation.
You may assume the dictionary does not contain duplicate words.
Example 1:

Input:
s = “catsanddog”
wordDict = [“cat”, “cats”, “and”, “sand”, “dog”]
Output:
[
“cats and dog”,
“cat sand dog”
]
Example 2:

Input:
s = “pineapplepenapple”
wordDict = [“apple”, “pen”, “applepen”, “pine”, “pineapple”]
Output:
[
“pine apple pen apple”,
“pineapple pen apple”,
“pine applepen apple”
]
Explanation: Note that you are allowed to reuse a dictionary word.
Example 3:

Input:
s = “catsandog”
wordDict = [“cats”, “dog”, “sand”, “and”, “cat”]
Output:
[]

TLE:

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class Solution:
def wordBreak(self, s: str, wordDict: List[str]) -> List[str]:
ans = []
def dfs(cur, idx):
if idx == len(s):
ans.append(' '.join(cur[:]))
for i in range(idx, len(s)):
w = s[idx:i + 1]
if w in wordDict:
cur.append(w)
dfs(cur, i + 1)
del cur[-1]
dfs([], 0)
return ans

AC: dfs + memo

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class Solution:
def wordBreak(self, s: str, wordDict: List[str]) -> List[str]:
def dfs(idx):
if idx == len(s):
return [""]
if idx in memo:
return memo[idx]
res = []
for i in range(idx, len(s)):
prefix = s[idx:i + 1]
if prefix in wordDict:
if i == len(s) - 1:
res.append(prefix)
else:
restW = dfs(i + 1)
for item in restW:
sentence = prefix + ' ' + item
res.append(sentence)
memo[idx] = res
return res
memo = {}
return dfs(0)

三刷:面经:Amazon。九章

Permutation based DFS - 基于排列的深度优先搜索

46. Permutations (Medium)

Given a collection of distinct integers, return all possible permutations.

Example:
Input: [1,2,3]
Output:
[
[1,2,3],
[1,3,2],
[2,1,3],
[2,3,1],
[3,1,2],
[3,2,1]
]

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class Solution:
def permute(self, nums):
ans = []
used = [False] * len(nums)
def dfs(cur):
if len(cur) == len(nums):
ans.append(cur[:])
return
for i in range(len(nums)):
if not used[i]:
used[i] = True
cur.append(nums[i])
dfs(cur)
del cur[-1]
used[i] = False
dfs([])
return ans

七刷:高频, subset从idx开始往后扫,全排每一位都有可能放任意数字,因此要从第一位往后扫。这样就需要一个used的变量来记住哪一位已经被用过了

47. Permutations II (Medium)

Given a collection of numbers that might contain duplicates, return all possible unique permutations.

Example:
Input: [1,1,2]
Output:
[
[1,1,2],
[1,2,1],
[2,1,1]
]

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class Solution:
def permuteUnique(self, nums):
ans = []
used = [False] * len(nums)
nums.sort()
def dfs(cur):
if len(cur) == len(nums):
ans.append(cur[:])
return
for i in range(len(nums)):
if i > 0 and nums[i] == nums[i -1] and not used[i - 1]:
continue
if not used[i]:
used[i] = True
cur.append(nums[i])
dfs(cur)
del cur[-1]
used[i] = False
dfs([])
return ans

六刷:高频,关键:when a number has the same value with its previous, we can use this number only if his previous is used

22. Generate Parentheses (Medium)

Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.

For example, given n = 3, a solution set is:
[
“((()))”,
“(()())”,
“(())()”,
“()(())”,
“()()()”
]

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class Solution:
def generateParenthesis(self, n):
ans = []
def dfs(cur, l, r):
if len(cur) == 2 * n:
ans.append(cur)
return
if l < n:
cur += '('
dfs(cur, l + 1, r)
cur = cur[:-1]
if r < l:
cur += ')'
dfs(cur, l, r + 1)
cur = cur[:-1]
dfs('', 0, 0)
return ans

五刷:高频

51. N-Queens (Hard)

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The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.
Given an integer n, return all distinct solutions to the n-queens puzzle.
Each solution contains a distinct board configuration of the n-queens' placement, where 'Q' and '.' both indicate a queen and an empty space respectively.

Example:

Input: 4
Output: [
[".Q..", // Solution 1
"...Q",
"Q...",
"..Q."],

["..Q.", // Solution 2
"Q...",
"...Q",
".Q.."]
]
Explanation: There exist two distinct solutions to the 4-queens puzzle as shown above.
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class Solution:
def solveNQueens(self, n: int) -> List[List[str]]:
res = []
def dfs(cur):
if len(cur) == n:
res.append(cur[:])
return
for c in range(n):
if legal(cur, c):
cur.append(c)
dfs(cur)
del cur[-1]
def legal(cur, c):
if c in cur:
return False
for cQ, rQ in enumerate(cur):
if cQ + rQ == len(cur) + c:
return False
if cQ - rQ == len(cur) - c:
return False
return True
dfs([])
return [["." * i + "Q" + "." * (n - i - 1) for i in b] for b in res]

二刷:高频。检查下一行的某列的合法性:举个栗子就能看出来,检查列就看此列是否已在cur中(皇后已在此列),检查/方向的对角线是坐标相加,检查\方向的对角线是坐标相减,行递增无需检查。时间:O(n^3), 空间:O(n)

Graph based DFS 基于图的深度优先搜索

17. Letter Combinations of a Phone Number (Medium)

Given a string containing digits from 2-9 inclusive, return all possible letter combinations that the number could represent.
A mapping of digit to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
keypad example
Example:
Input: “23”
Output: [“ad”, “ae”, “af”, “bd”, “be”, “bf”, “cd”, “ce”, “cf”].

Note:
Although the above answer is in lexicographical order, your answer could be in any order you want.

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class Solution:
def letterCombinations(self, digits: str) -> List[str]:
if not digits:
return []
cMap = {"2": "abc", "3": "def", "4": "ghi", "5": "jkl", "6": "mno", "7": "pqrs", "8": "tuv", "9": "wxyz"}
ans = []
def dfs(cur, idx):
if len(cur) == len(digits):
ans.append("".join(cur))
return
for c in cMap[digits[idx]]:
cur.append(c)
dfs(cur, idx + 1)
del cur[-1]
dfs([], 0)
return ans]

三刷:高频

79. Word Search (Medium)

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Given a 2D board and a word, find if the word exists in the grid.

The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.

Example:

board =
[
['A','B','C','E'],
['S','F','C','S'],
['A','D','E','E']
]

Given word = "ABCCED", return true.
Given word = "SEE", return true.
Given word = "ABCB", return false.
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class Solution:
def exist(self, board: List[List[str]], word: str) -> bool:
ans = False
def dfs(r, c, idx):
nonlocal ans
if idx == len(word):
ans = True
return
if not ans and legal(r, c) and board[r][c] == word[idx]:
tmp = board[r][c]
board[r][c] = '#'
dfs(r - 1, c, idx + 1)
dfs(r + 1, c, idx + 1)
dfs(r, c - 1, idx + 1)
dfs(r, c + 1, idx + 1)
if not ans:
board[r][c] = tmp
def legal(r, c):
return r >= 0 and c >= 0 and r < len(board) and c < len(board[0])
for r in range(len(board)):
for c in range(len(board[0])):
dfs(r, c, 0)
return ans

二刷:高频

490. The Maze (Medium)

带锁题
There is a ball in a maze with empty spaces and walls. The ball can go through empty spaces by rolling up, down, left or right, but it won’t stop rolling until hitting a wall. When the ball stops, it could choose the next direction.

Given the ball’s start position, the destination and the maze, determine whether the ball could stop at the destination.

The maze is represented by a binary 2D array. 1 means the wall and 0 means the empty space. You may assume that the borders of the maze are all walls. The start and destination coordinates are represented by row and column indexes.

Example 1
Input 1: a maze represented by a 2D array
0 0 1 0 0
0 0 0 0 0
0 0 0 1 0
1 1 0 1 1
0 0 0 0 0
Input 2: start coordinate (rowStart, colStart) = (0, 4)
Input 3: destination coordinate (rowDest, colDest) = (4, 4)

Output: true
Explanation: One possible way is : left -> down -> left -> down -> right -> down -> right.
the maze example 1

Example 2
Input 1: a maze represented by a 2D array
0 0 1 0 0
0 0 0 0 0
0 0 0 1 0
1 1 0 1 1
0 0 0 0 0
Input 2: start coordinate (rowStart, colStart) = (0, 4)
Input 3: destination coordinate (rowDest, colDest) = (3, 2)

Output: false
Explanation: There is no way for the ball to stop at the destination.
the maze example 2

Note:
1.There is only one ball and one destination in the maze.
2.Both the ball and the destination exist on an empty space, and they will not be at the same position initially.
3.The given maze does not contain border (like the red rectangle in the example pictures), but you could assume the border of the maze are all walls.
4.The maze contains at least 2 empty spaces, and both the width and height of the maze won’t exceed 100.

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class Solution(object):
def hasPath(self, maze, start, destination):
"""
:type maze: List[List[int]]
:type start: List[int]
:type destination: List[int]
:rtype: bool
"""
visited = set()
m, n = len(maze), len(maze[0])
return self.dfs(maze, (start[0], start[1]), (destination[0], destination[1]), visited, m, n)
def dfs(self, maze, cur, dest, visited, m, n):
if cur == dest:
return True
if cur in visited:
return False
visited.add(cur)
for rd, cd in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
r, c = cur
while 0 <= r + rd <= m - 1 and 0 <= c + cd <= n - 1 and maze[r + rd][c + cd] != 1:
r += rd
c += cd
if self.dfs(maze, (r, c), dest, visited, m, n):
return True
return False

面经:Amazon。注意…r, c = cur…必须在for下面,if self.dfs(…): return True…必须是这个结构。todo: BFS

417. Pacific Atlantic Water Flow (Medium)

Given an m x n matrix of non-negative integers representing the height of each unit cell in a continent, the “Pacific ocean” touches the left and top edges of the matrix and the “Atlantic ocean” touches the right and bottom edges.
Water can only flow in four directions (up, down, left, or right) from a cell to another one with height equal or lower.
Find the list of grid coordinates where water can flow to both the Pacific and Atlantic ocean.

Note:
The order of returned grid coordinates does not matter.
Both m and n are less than 150.

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Example:
Given the following 5x5 matrix:
Pacific ~ ~ ~ ~ ~
~ 1 2 2 3 (5) *
~ 3 2 3 (4) (4) *
~ 2 4 (5) 3 1 *
~ (6) (7) 1 4 5 *
~ (5) 1 1 2 4 *
* * * * * Atlantic
Return:
[[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]] (positions with parentheses in above matrix).
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class Solution:
def pacificAtlantic(self, matrix: List[List[int]]) -> List[List[int]]:
if not matrix:
return []
m, n = len(matrix), len(matrix[0])
reach_p = [[False] * n for _ in range(m)]
reach_a = [[False] * n for _ in range(m)]
self.dirs = [(-1, 0), (1, 0), (0, 1), (0, -1)]
for r in range(m):
self.dfs(matrix, r, 0, m, n, reach_p)
self.dfs(matrix, r, n - 1, m, n, reach_a)
for c in range(n):
self.dfs(matrix, 0, c, m, n, reach_p)
self.dfs(matrix, m - 1, c, m, n, reach_a)
return [[r, c] for r in range(m) for c in range(n) if reach_p[r][c] == True and reach_a[r][c] == True]

def dfs(self, matrix, r, c, m, n, reach):
reach[r][c] = True
for rd, cd in self.dirs:
nr, nc = r + rd, c + cd
if 0 <= nr <= m - 1 and 0 <= nc <= n - 1 and matrix[nr][nc] >= matrix[r][c] and not reach[nr][nc]:

self.dfs(matrix, nr, nc, m, n, reach)

面经:Cruise Automation。dfs:从两个海开始往内陆灌,返回两个海都能灌到的地方 todo bfs

399. Evaluate Division (Medium)

Equations are given in the format A / B = k, where A and B are variables represented as strings, and k is a real number (floating point number). Given some queries, return the answers. If the answer does not exist, return -1.0.

Example:
Given a / b = 2.0, b / c = 3.0.
queries are: a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ? .
return [6.0, 0.5, -1.0, 1.0, -1.0 ].

The input is: vector<pair<string, string>> equations, vector& values, vector<pair<string, string>> queries , where equations.size() == values.size(), and the values are positive. This represents the equations. Return vector.

According to the example above:
equations = [ [“a”, “b”], [“b”, “c”] ],
values = [2.0, 3.0],
queries = [ [“a”, “c”], [“b”, “a”], [“a”, “e”], [“a”, “a”], [“x”, “x”] ].

The input is always valid. You may assume that evaluating the queries will result in no division by zero and there is no contradiction.

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class Solution:
def calcEquation(self, equations: List[List[str]], values: List[float], queries: List[List[str]]) -> List[float]:
from collections import defaultdict
graph = defaultdict(list)
self.buildGraph(graph, equations, values)
return [self.findPath(graph, s, e) for s, e in queries]
def buildGraph(self, graph, equations, values):
for vertices, value in zip(equations, values):
a, b = vertices
graph[a].append((b, value))
graph[b].append((a, 1 / value))
def findPath(self, graph, s, e):
if s not in graph:
return -1.0
q = [(s, 1.0)]
visited = set()
while q:
cur, product = q.pop(0)
if cur == e:
return product
visited.add(cur)
for vertex, value in graph[cur]:
if vertex not in visited:
q.append((vertex, value * product))
return -1.0

面经:头条。Currency Calculator一种题,先build graph,再用BFS寻最短路径的乘积

332. Reconstruct Itinerary (Medium)

Given a list of airline tickets represented by pairs of departure and arrival airports [from, to], reconstruct the itinerary in order. All of the tickets belong to a man who departs from JFK. Thus, the itinerary must begin with JFK.

Note:
If there are multiple valid itineraries, you should return the itinerary that has the smallest lexical order when read as a single string. For example, the itinerary [“JFK”, “LGA”] has a smaller lexical order than [“JFK”, “LGB”].
All airports are represented by three capital letters (IATA code).
You may assume all tickets form at least one valid itinerary.

Example 1:
Input: [[“MUC”, “LHR”], [“JFK”, “MUC”], [“SFO”, “SJC”], [“LHR”, “SFO”]]
Output: [“JFK”, “MUC”, “LHR”, “SFO”, “SJC”]

Example 2:
Input: [[“JFK”,”SFO”],[“JFK”,”ATL”],[“SFO”,”ATL”],[“ATL”,”JFK”],[“ATL”,”SFO”]]
Output: [“JFK”,”ATL”,”JFK”,”SFO”,”ATL”,”SFO”]
Explanation: Another possible reconstruction is [“JFK”,”SFO”,”ATL”,”JFK”,”ATL”,”SFO”].
But it is larger in lexical order.

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class Solution:
def findItinerary(self, tickets: List[List[str]]) -> List[str]:
from collections import defaultdict
flights = defaultdict(list)
for t in tickets:
flights[t[0]].append(t[1])
for k in flights:
flights[k].sort()
ans = ["JFK"]
self.dfs("JFK", flights, ans, len(tickets))
return ans
def dfs(self, departure, flights, path, numTix):
if departure not in flights:
return
arrival = flights[departure]
for i, a in enumerate(arrival):
path.append(a)
del arrival[i]
self.dfs(a, flights, path, numTix)
if len(path) == numTix + 1:
return
arrival.insert(i, a)
del path[-1]

一刷:这个if len(path) == numTix + 1必须在递归调用后面来查看是否已满足需求,即不剪枝 todo:没有100%的理解

数据结构

Array 数组

LinC 6. Merge Two Sorted Arrays (Easy)

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Merge two given sorted integer array A and B into a new sorted integer array.

Example
A=[1,2,3,4]

B=[2,4,5,6]

return [1,2,2,3,4,4,5,6]

Challenge
How can you optimize your algorithm if one array is very large and the other is very small?

思路:热身题,直接做

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class Solution:
"""
@param A: sorted integer array A
@param B: sorted integer array B
@return: A new sorted integer array
"""
def mergeSortedArray(self, A, B):
# write your code here
ans = []
indexA = 0
indexB = 0
indexC = 0
while indexC < len(A) + len(B):
if indexA == len(A) or indexB == len(B):
if indexA == len(A):
ans.append(B[indexB])
indexB += 1
else:
ans.append(A[indexA])
indexA += 1
else:
if A[indexA] < B[indexB]:
ans.append(A[indexA])
indexA += 1
else:
ans.append(B[indexB])
indexB += 1
indexC += 1
return ans

总结:非常值得刷的一道热身题, 需要考虑两个 array 越界的问题。看了下答案用三个 while 循环也可以。

88. Merge Sorted Array (Easy)

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Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.

Note:

The number of elements initialized in nums1 and nums2 are m and n respectively.
You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2.
Example:

Input:
nums1 = [1,2,3,0,0,0], m = 3
nums2 = [2,5,6], n = 3

Output: [1,2,2,3,5,6]

思路:直觉上想不太出来怎么不创建新的存储空间把小数组 merge 到大数组里。看了答案,如果 nums1 后面空着这么些空,就从后面开始填。哎,曾经是能自主想的出的。。。正着困难的话就反着试试

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class Solution(object):
def merge(self, nums1, m, nums2, n):
"""
:type nums1: List[int]
:type m: int
:type nums2: List[int]
:type n: int
:rtype: void Do not return anything, modify nums1 in-place instead.
"""
if len(nums2) == 0:
return
if len(nums1) == len(nums2):
for i in range(len(nums2)):
nums1[i] = nums2[i]
return
index1 = m - 1
index2 = n - 1
index3 = len(nums1) - 1
while index3 >= 0 and index2 >= 0:
if index1 < 0:
nums1[index3] = nums2[index2]
index2 -= 1
else:
if nums1[index1] < nums2[index2]:
nums1[index3] = nums2[index2]
index2 -= 1
else:
nums1[index3] = nums1[index1]
index1 -= 1
index3 -= 1

总结:虽然是 easy 题,要考虑情况:1.nums1 和 nums2 一样大的话需要逐个考过去;2.index2 如果走到最前面就可以结束了。注意题目的输入包含了 m 和 n 要利用好

高频:

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class Solution:
def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
"""
Do not return anything, modify nums1 in-place instead.
"""
if m == 0:
nums1[:] = nums2[:]
i1 = m - 1
i2 = n - 1
im = m + n - 1
while i1 >= 0 and i2 >= 0:
if nums1[i1] < nums2[i2]:
nums1[im] = nums2[i2]
i2 -= 1
else:
nums1[im] = nums1[i1]
i1 -= 1
im -= 1
if i1 < 0:
nums1[:i2 + 1] = nums2[:i2 + 1]

总结:代码简化,也更好理解一些,算法还是原来的,从nums1后往前填��注意最后如果nums1都填完了要把nums2剩余的都天过去nums1[:i2 + 1] = nums2[:i2 + 1]

73. Set Matrix Zeroes (Medium)

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Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in-place.

Example 1:

Input:
[
[1,1,1],
[1,0,1],
[1,1,1]
]
Output:
[
[1,0,1],
[0,0,0],
[1,0,1]
]
Example 2:

Input:
[
[0,1,2,0],
[3,4,5,2],
[1,3,1,5]
]
Output:
[
[0,0,0,0],
[0,4,5,0],
[0,3,1,0]
]
Follow up:

A straight forward solution using O(mn) space is probably a bad idea.
A simple improvement uses O(m + n) space, but still not the best solution.
Could you devise a constant space solution?

高频

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class Solution:
def setZeroes(self, matrix: List[List[int]]) -> None:
"""
Do not return anything, modify matrix in-place instead.
"""
r0 = set()
c0 = set()

for r in range(len(matrix)):
for c in range(len(matrix[0])):
if matrix[r][c] == 0:
r0.add(r)
c0.add(c)

for r in r0:
for c in range(len(matrix[0])):
matrix[r][c] = 0

for c in c0:
for r in range(len(matrix)):
matrix[r][c] = 0

总结:O(m+n)空间的解法
二刷:

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class Solution:
def setZeroes(self, matrix: List[List[int]]) -> None:
is1stRZ, is1stCZ = True if 0 in matrix[0] else False, True if 0 in list(zip(*matrix))[0] else False
for i, r in enumerate(matrix):
for j, c in enumerate(r):
if i and j and c == 0:
matrix[i][0] = 0
matrix[0][j] = 0
m, n = len(matrix), len(matrix[0])
for i in range(1, m):
if matrix[i][0] == 0:
matrix[i] = [0] * n
for j in range(1, n):
if matrix[0][j] == 0:
for i in range(1, m):
matrix[i][j] = 0
if is1stRZ:
matrix[0] = [0] * n
if is1stCZ:
for i in range(m):
matrix[i][0] = 0

总结:O(1)空间

56. Merge Intervals (Medium)

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Given a collection of intervals, merge all overlapping intervals.

Example 1:

Input: [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].
Example 2:

Input: [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.

高频

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class Solution:
def merge(self, intervals: List[List[int]]) -> List[List[int]]:
if not intervals:
return []
intervals.sort(key = lambda x: x[0])
ans = []
cur = intervals[0]
for i in range(1, len(intervals)):
if intervals[i][0] <= cur[1]:
cur[1] = max(intervals[i][1], cur[1])
else:
ans.append(cur[:])
cur = intervals[i]
ans.append(cur)
return ans

面经:Cruise。…cur[1] = max(intervals[i][1], cur[1])

LinC 839. Merge Two Sorted Interval Lists (Easy)

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Merge two sorted (ascending) lists of interval and return it as a new sorted list. The new sorted list should be made by splicing together the intervals of the two lists and sorted in ascending order.

The intervals in the given list do not overlap.
The intervals in different lists may overlap.
Example
Given list1 = [(1,2),(3,4)] and list2 = [(2,3),(5,6)], return [(1,4),(5,6)].

思路:思路跟上题 merge interval一样,可以不做

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"""
Definition of Interval.
class Interval(object):
def __init__(self, start, end):
self.start = start
self.end = end
"""
class Solution:
"""
@param list1: one of the given list
@param list2: another list
@return: the new sorted list of interval
"""
def mergeTwoInterval(self, list1, list2):
# write your code here
if not list1:
return list2
if not list2:
return list1
list3 = list1 + list2
list3.sort(key=lambda x: x.start)
ans = [list3[0]]
for i in range(1, len(list3)):
if list3[i].start <= ans[-1].end:
ans[-1].end = max(list3[i].end, ans[-1].end)
else:
ans.append(list3[i])
return ans

二刷:删掉一刷代码,统一思路

228. Summary Ranges (Medium)

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Given a sorted integer array without duplicates, return the summary of its ranges.

Example 1:

Input: [0,1,2,4,5,7]
Output: ["0->2","4->5","7"]
Explanation: 0,1,2 form a continuous range; 4,5 form a continuous range.
Example 2:

Input: [0,2,3,4,6,8,9]
Output: ["0","2->4","6","8->9"]
Explanation: 2,3,4 form a continuous range; 8,9 form a continuous range.
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# 先写 O(n) 的再优化
class Solution:
def summaryRanges(self, nums):
"""
:type nums: List[int]
:rtype: List[str]
"""
if not nums:
return []
ans, start = [], 0
def addToRes(l, r):
if l == r:
ans.append(str(nums[l]))
else:
ans.append(f"{str(nums[l])}->{str(nums[r])}")
for i in range(len(nums) - 1):
if nums[i] != nums[i + 1] - 1:
addToRes(start, i)
start = i + 1
addToRes(start, len(nums) - 1)
return ans

二刷:删掉了一刷的思路,代码和总结。 ..start = 0…def addToRes(l, r):…start = i + 1…

67. Add Binary (Easy)

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Given two binary strings, return their sum (also a binary string).

The input strings are both non-empty and contains only characters 1 or 0.

Example 1:

Input: a = "11", b = "1"
Output: "100"
Example 2:

Input: a = "1010", b = "1011"
Output: "10101"

高频

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class Solution:
def addBinary(self, a: str, b: str) -> str:
m = len(a)
n = len(b)
l = max(m, n)
i = 1
carry = 0
ans = ""
while i <= l:
if i <= m and i <= n:
val = int(a[-i]) + int(b[-i]) + carry
elif i <= m:
val = int(a[-i]) + carry
elif i <= n:
val = int(b[-i]) + carry
if val > 1:
carry = 1
else:
carry = 0
ans = str(val % 2) + ans
i += 1
if carry == 1:
ans = "1" + ans
return ans

总结:…i = 1…while i <= l:…if val > 1: carry = 1; else: carry = 0; ans = str(val % 2) + ans; i += 1…

12. Integer to Roman (Medium)

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Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

Symbol Value
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
For example, two is written as II in Roman numeral, just two one's added together. Twelve is written as, XII, which is simply X + II. The number twenty seven is written as XXVII, which is XX + V + II.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:

I can be placed before V (5) and X (10) to make 4 and 9.
X can be placed before L (50) and C (100) to make 40 and 90.
C can be placed before D (500) and M (1000) to make 400 and 900.
Given an integer, convert it to a roman numeral. Input is guaranteed to be within the range from 1 to 3999.

Example 1:

Input: 3
Output: "III"
Example 2:

Input: 4
Output: "IV"
Example 3:

Input: 9
Output: "IX"
Example 4:

Input: 58
Output: "LVIII"
Explanation: L = 50, V = 5, III = 3.
Example 5:

Input: 1994
Output: "MCMXCIV"
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

高频

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class Solution:
def intToRoman(self, num: int) -> str:
n = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1]
l = ["M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"]
ans = ""
for i, v in enumerate(n):
if num == 0:
return ans
t = num // v
ans += l[i] * t
num %= v
return ans

总结:有了n和l俩数组就是easy了

43. Multiply Strings (Medium)

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Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2, also represented as a string.

Example 1:

Input: num1 = "2", num2 = "3"
Output: "6"
Example 2:

Input: num1 = "123", num2 = "456"
Output: "56088"
Note:

The length of both num1 and num2 is < 110.
Both num1 and num2 contain only digits 0-9.
Both num1 and num2 do not contain any leading zero, except the number 0 itself.
You must not use any built-in BigInteger library or convert the inputs to integer directly.
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class Solution:
def multiply(self, num1: str, num2: str) -> str:
res = [0 for _ in range(len(num1) + len(num2))]
num1 = num1[::-1]
num2 = num2[::-1]
for i in range(len(num1)):
for j in range(len(num2)):
res[i + j] += int(num1[i]) * int(num2[j])

for i in range(len(res)):
d = res[i] % 10
c = res[i] // 10
if i < len(res) - 1:
res[i + 1] += c
res[i] = d

res.reverse()
while res[0] == 0 and len(res) > 1:
del res[0]
return "".join(str(i) for i in res)

高频:…for i in range(len(res)):…res[i] = d…while res[0] == 0 and len(res) > 1:…

128. Longest Consecutive Sequence (Medium)

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Given an unsorted array of integers, find the length of the longest consecutive elements sequence.

Your algorithm should run in O(n) complexity.

Example:

Input: [100, 4, 200, 1, 3, 2]
Output: 4
Explanation: The longest consecutive elements sequence is [1, 2, 3, 4]. Therefore its length is 4.
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class Solution:
def longestConsecutive(self, nums: List[int]) -> int:
nums = set(nums)
ans = 0
for n in nums:
if n - 1 not in nums:
m = n + 1
while m in nums:
m += 1
ans = max(ans, m - n)
return ans

高频:不明白为什么要作为高频题,除了惊叹于答案之神乎其技, 几乎学不到任何东西

66. Plus One (Easy)

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Given a non-empty array of digits representing a non-negative integer, plus one to the integer.

The digits are stored such that the most significant digit is at the head of the list, and each element in the array contain a single digit.

You may assume the integer does not contain any leading zero, except the number 0 itself.

Example 1:

Input: [1,2,3]
Output: [1,2,4]
Explanation: The array represents the integer 123.
Example 2:

Input: [4,3,2,1]
Output: [4,3,2,2]
Explanation: The array represents the integer 4321.
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class Solution:
def plusOne(self, digits: List[int]) -> List[int]:
digits.reverse()
c = 0
for i, d in enumerate(digits):
if i == 0:
d += 1
else:
d += c
c = d // 10
d = d % 10
digits[i] = d
if c > 0:
digits.append(1)
return digits[::-1]

高频

9. Palindrome Number (Easy)

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Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.

Example 1:

Input: 121
Output: true
Example 2:

Input: -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.
Example 3:

Input: 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.
Follow up:

Coud you solve it without converting the integer to a string?
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class Solution:
def isPalindrome(self, x: int) -> bool:
if x < 0:
return False
ox = x
nx = 0
while x:
t = x % 10
nx = nx * 10 + t
x //= 10
return nx == ox

高频:…nx = nx * 10 + t…

59. Spiral Matrix II (Medium)

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Given a positive integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.

Example:

Input: 3
Output:
[
[ 1, 2, 3 ],
[ 8, 9, 4 ],
[ 7, 6, 5 ]
]
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class Solution:
def generateMatrix(self, n: int) -> List[List[int]]:
ans = [[0] * n for _ in range(n)]
mode, r, c, circle = 0, 0, 0, 0
for i in range(1, n * n + 1):
ans[r][c] = i
if mode == 0:
c += 1
if c == n - 1 - circle:
mode = 1
elif mode == 1:
r += 1
if r == n - 1 - circle:
mode = 2
elif mode == 2:
c -= 1
if c == circle:
mode = 3
else:
r -= 1
if r == circle + 1:
mode = 0
circle += 1
return ans

高频:…if c == n - 1 - circle:…if r == n - 1 - circle:…if c == circle:…if r = circle + 1:…

48. Rotate Image (Medium)

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You are given an n x n 2D matrix representing an image.
Rotate the image by 90 degrees (clockwise).
Note:
You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation.

Example 1:
Given input matrix =
[
[1,2,3],
[4,5,6],
[7,8,9]
],

rotate the input matrix in-place such that it becomes:
[
[7,4,1],
[8,5,2],
[9,6,3]
]
Example 2:

Given input matrix =
[
[ 5, 1, 9,11],
[ 2, 4, 8,10],
[13, 3, 6, 7],
[15,14,12,16]
],

rotate the input matrix in-place such that it becomes:
[
[15,13, 2, 5],
[14, 3, 4, 1],
[12, 6, 8, 9],
[16, 7,10,11]
]
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class Solution:
def rotate(self, matrix: List[List[int]]) -> None:
"""
Do not return anything, modify matrix in-place instead.
"""
n = len(matrix)
matrix.reverse()
for r in range(n):
for c in range(r, n):
matrix[r][c], matrix[c][r] = matrix[c][r], matrix[r][c]

高频:需要懂得顺时针转就是上下翻转以后对角线翻转,逆时针转就是左右翻转以后对角线翻转。懂list(zip(*list))的话就是一行代码

54. Spiral Matrix (Medium)

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Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

Example 1:

Input:
[
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]
]
Output: [1,2,3,6,9,8,7,4,5]
Example 2:

Input:
[
[1, 2, 3, 4],
[5, 6, 7, 8],
[9,10,11,12]
]
Output: [1,2,3,4,8,12,11,10,9,5,6,7]
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class Solution:
def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
if len(matrix) == 0:
return []
mode, circle, r, c = 0, 0, 0, 0
m, n = len(matrix), len(matrix[0])
ans = []
for _ in range(m * n):
ans.append(matrix[r][c])
if mode == 0:
if c == n - 1 - circle:
mode = 1
r += 1
else:
c += 1
elif mode == 1:
if r == m - 1 - circle:
mode = 2
c -= 1
else:
r += 1
elif mode == 2:
if c == circle:
mode = 3
r -= 1
else:
c -= 1
elif mode == 3:
if r == circle + 1:
mode = 0
circle += 1
c += 1
else:
r -= 1
return ans

高频:相比其他答案,更喜欢这个sprial matrix II用过的模板,不同之处是这里要先处理转向,用来应对[[1],[2]]这种输入

68. Text Justification (Hard)

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Given an array of words and a width maxWidth, format the text such that each line has exactly maxWidth characters and is fully (left and right) justified.

You should pack your words in a greedy approach; that is, pack as many words as you can in each line. Pad extra spaces ' ' when necessary so that each line has exactly maxWidth characters.

Extra spaces between words should be distributed as evenly as possible. If the number of spaces on a line do not divide evenly between words, the empty slots on the left will be assigned more spaces than the slots on the right.

For the last line of text, it should be left justified and no extra space is inserted between words.

Note:

A word is defined as a character sequence consisting of non-space characters only.
Each word's length is guaranteed to be greater than 0 and not exceed maxWidth.
The input array words contains at least one word.
Example 1:

Input:
words = ["This", "is", "an", "example", "of", "text", "justification."]
maxWidth = 16
Output:
[
"This is an",
"example of text",
"justification. "
]
Example 2:

Input:
words = ["What","must","be","acknowledgment","shall","be"]
maxWidth = 16
Output:
[
"What must be",
"acknowledgment ",
"shall be "
]
Explanation: Note that the last line is "shall be " instead of "shall be",
because the last line must be left-justified instead of fully-justified.
Note that the second line is also left-justified becase it contains only one word.
Example 3:

Input:
words = ["Science","is","what","we","understand","well","enough","to","explain",
"to","a","computer.","Art","is","everything","else","we","do"]
maxWidth = 20
Output:
[
"Science is what we",
"understand well",
"enough to explain to",
"a computer. Art is",
"everything else we",
"do "
]
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class Solution:
def fullJustify(self, words: List[str], maxWidth: int) -> List[str]:
ans, line, letter_count = [], [], 0
for w in words:
if letter_count + len(w) + len(line) > maxWidth:
for i in range(maxWidth - letter_count):
line[i % (len(line) - 1 or 1)] += ' '
ans.append(''.join(line))
line = []
letter_count = 0
line += [w]
letter_count += len(w)
ans.append(' '.join(line).ljust(maxWidth))
return ans

高频:需要知道很巧妙的round robin填空格的方法:…line[i % (len(line) - 1 or 1)] += ‘ ‘…,str.ljust(width[, fillchar])默认填空格

36. Valid Sudoku (Medium)

Determine if a 9x9 Sudoku board is valid. Only the filled cells need to be validated according to the following rules:
Each row must contain the digits 1-9 without repetition.
Each column must contain the digits 1-9 without repetition.
Each of the 9 3x3 sub-boxes of the grid must contain the digits 1-9 without repetition.
A partially filled sudoku which is valid.
The Sudoku board could be partially filled, where empty cells are filled with the character ‘.’.

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Example 1:
Input:
[
["5","3",".",".","7",".",".",".","."],
["6",".",".","1","9","5",".",".","."],
[".","9","8",".",".",".",".","6","."],
["8",".",".",".","6",".",".",".","3"],
["4",".",".","8",".","3",".",".","1"],
["7",".",".",".","2",".",".",".","6"],
[".","6",".",".",".",".","2","8","."],
[".",".",".","4","1","9",".",".","5"],
[".",".",".",".","8",".",".","7","9"]
]
Output: true

Example 2:
Input:
[
["8","3",".",".","7",".",".",".","."],
["6",".",".","1","9","5",".",".","."],
[".","9","8",".",".",".",".","6","."],
["8",".",".",".","6",".",".",".","3"],
["4",".",".","8",".","3",".",".","1"],
["7",".",".",".","2",".",".",".","6"],
[".","6",".",".",".",".","2","8","."],
[".",".",".","4","1","9",".",".","5"],
[".",".",".",".","8",".",".","7","9"]
]
Output: false
Explanation: Same as Example 1, except with the 5 in the top left corner being
modified to 8. Since there are two 8's in the top left 3x3 sub-box, it is invalid.

Note:
A Sudoku board (partially filled) could be valid but is not necessarily solvable.
Only the filled cells need to be validated according to the mentioned rules.
The given board contain only digits 1-9 and the character ‘.’.
The given board size is always 9x9.

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class Solution:
def isValidSudoku(self, board: List[List[str]]) -> bool:
#check every row
for r in board:
if not self.validate(r):
return False
#check every col
for c in zip(*board):
if not self.validate(list(c)):
return False
#check every 3x3
for dr in [0, 3, 6]:
for dc in [0, 3, 6]:
flat = []
for r in range(0, 3):
for c in range(0, 3):
flat.append(board[r + dr][c + dc])
if not self.validate(flat):
return False
return True
def validate(self, r):
visited = set()
for c in r:
if c != ".":
if c in visited:
return False
else:
visited.add(c)
return True

高频, 面经:维萨。*用来unpack二维数组,zip(*board)用来返回一个用每行同一列位置的元素拼成的新的tuple的iterator

38. Count and Say (Easy)

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The count-and-say sequence is the sequence of integers with the first five terms as following:

1. 1
2. 11
3. 21
4. 1211
5. 111221
1 is read off as "one 1" or 11.
11 is read off as "two 1s" or 21.
21 is read off as "one 2, then one 1" or 1211.

Given an integer n where 1 ≤ n ≤ 30, generate the nth term of the count-and-say sequence.
Note: Each term of the sequence of integers will be represented as a string.

Example 1:

Input: 1
Output: "1"
Example 2:

Input: 4
Output: "1211"

用itertools.groupby():

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class Solution:
def countAndSay(self, n: int) -> str:
ans = "1"
for i in range(n - 1):
t = ""
for d, g in itertools.groupby(ans):
cnt = len(list(g))
t += str(cnt) + str(d)
ans = t
return ans

不用groupby:

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class Solution:
def countAndSay(self, n: int) -> str:
ans = "1"
for i in range(n - 1):
res = ""
t = ans + "#"
cnt = 1
for i in range(len(t) - 1):
if t[i] == t[i + 1]:
cnt += 1
else:
res += str(cnt) + t[i]
cnt = 1
ans = res
return ans

高频:题目描述不是很清楚,用groupby:…for d, g in itertools.groupby():…。不用groupby:…res = “”; t = ans + “#”…

14. Longest Common Prefix (Easy)

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Write a function to find the longest common prefix string amongst an array of strings.

If there is no common prefix, return an empty string "".

Example 1:

Input: ["flower","flow","flight"]
Output: "fl"
Example 2:

Input: ["dog","racecar","car"]
Output: ""
Explanation: There is no common prefix among the input strings.
Note:

All given inputs are in lowercase letters a-z.
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class Solution:
def longestCommonPrefix(self, strs: List[str]) -> str:
if not strs:
return ""
ans = strs[0]
for i in range(1, len(strs)):
if not ans:
return ""
else:
j = 0
while j < len(ans) and j < len(strs[i]) and ans[j] == strs[i][j]:
j += 1
ans = ans[:j]
return ans

高频:
面经:Quora。删掉了高频的总结,换成我自己想出来的解法。

119. Pascal’s Triangle II (Easy)

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Given a non-negative index k where k ≤ 33, return the kth index row of the Pascal's triangle.
Note that the row index starts from 0.
In Pascal's triangle, each number is the sum of the two numbers directly above it.

Example:
Input: 3
Output: [1,3,3,1]

Follow up:
Could you optimize your algorithm to use only O(k) extra space?
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class Solution:
def getRow(self, rowIndex: int) -> List[int]:
ans = [1]
for i in range(rowIndex):
ans = [x + y for x, y in zip(ans + [0], [0] + ans)]
return ans

高频:很巧妙的…[x + y for x, y in zip(ans + [0], [0] + ans)]…

6. ZigZag Conversion (Medium)

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The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)

P A H N
A P L S I I G
Y I R
And then read line by line: "PAHNAPLSIIGYIR"

Write the code that will take a string and make this conversion given a number of rows:

string convert(string s, int numRows);
Example 1:

Input: s = "PAYPALISHIRING", numRows = 3
Output: "PAHNAPLSIIGYIR"
Example 2:

Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"
Explanation:

P I N
A L S I G
Y A H R
P I
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class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows == 1:
return s
res = [''] * numRows
r, move = 0, 1
for l in s:
res[r] += l
r += move
if r == numRows - 1:
move = -1
elif r == 0:
move = 1
return ''.join(res)

高频:想太多反而写不出来,直接粗暴拼反而能拼出来,注意numRows为1时单独处理

8. String to Integer (atoi) (Medium)

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Implement atoi which converts a string to an integer.

The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value.

The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function.

If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed.

If no valid conversion could be performed, a zero value is returned.

Note:

Only the space character ' ' is considered as whitespace character.
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. If the numerical value is out of the range of representable values, INT_MAX (231 − 1) or INT_MIN (−231) is returned.
Example 1:

Input: "42"
Output: 42
Example 2:

Input: " -42"
Output: -42
Explanation: The first non-whitespace character is '-', which is the minus sign.
Then take as many numerical digits as possible, which gets 42.
Example 3:

Input: "4193 with words"
Output: 4193
Explanation: Conversion stops at digit '3' as the next character is not a numerical digit.
Example 4:

Input: "words and 987"
Output: 0
Explanation: The first non-whitespace character is 'w', which is not a numerical
digit or a +/- sign. Therefore no valid conversion could be performed.
Example 5:

Input: "-91283472332"
Output: -2147483648
Explanation: The number "-91283472332" is out of the range of a 32-bit signed integer.
Thefore INT_MIN (−231) is returned.
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class Solution:
def myAtoi(self, str: str) -> int:
return self.helper(str)
def helper(self, s):
s = s.strip()
if not s:
return 0
valid = [str(i) for i in range(10)]
signs = '+-'
if s[0] not in valid and s[0] not in signs:
return 0
sign = 1
if s[0] == '-':
sign = -1
if s[0] in signs:
s = s[1:]
ans = ''
for l in s:
if l in valid:
ans += l
if sign == -1 and -int(ans) < -pow(2, 31):
return -pow(2, 31)
if sign == 1 and int(ans) > pow(2, 31) - 1:
return pow(2, 31) - 1
else:
return sign * int(ans) if ans else 0
return sign * int(ans) if ans else 0

高频:从例子之多就能看出来要考虑的case非常多。比的是细致

57. Insert Interval (Hard)

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Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).

You may assume that the intervals were initially sorted according to their start times.

Example 1:

Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
Output: [[1,5],[6,9]]
Example 2:

Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
Output: [[1,2],[3,10],[12,16]]
Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].
NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.
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class Solution:
def insert(self, intervals, newInterval):
if not intervals:
return [newInterval]
ans = []
i = 0
while i < len(intervals):
if intervals[i][1] < newInterval[0]:
ans.append(intervals[i])
i += 1
else:
break
while i < len(intervals) and intervals[i][0] <= newInterval[1]:
newInterval[0] = min(newInterval[0], intervals[i][0])
newInterval[1] = max(newInterval[1], intervals[i][1])
i += 1
ans.append(newInterval)
while i < len(intervals):
ans.append(intervals[i])
i += 1
return ans

高频:重点是知道先把,结束小于新开始的…if intervals[i][1] < newInterval[0]:…加到ans里,然后过开始小于新结束的…if…and intervals[i][0] <= newInterval[1]: …min…max…这个方法

937. Reorder Log Files (Easy)

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You have an array of logs.  Each log is a space delimited string of words.

For each log, the first word in each log is an alphanumeric identifier. Then, either:

Each word after the identifier will consist only of lowercase letters, or;
Each word after the identifier will consist only of digits.
We will call these two varieties of logs letter-logs and digit-logs. It is guaranteed that each log has at least one word after its identifier.

Reorder the logs so that all of the letter-logs come before any digit-log. The letter-logs are ordered lexicographically ignoring identifier, with the identifier used in case of ties. The digit-logs should be put in their original order.

Return the final order of the logs.
Example 1:

Input: ["a1 9 2 3 1","g1 act car","zo4 4 7","ab1 off key dog","a8 act zoo"]
Output: ["g1 act car","a8 act zoo","ab1 off key dog","a1 9 2 3 1","zo4 4 7"]

Note:

0 <= logs.length <= 100
3 <= logs[i].length <= 100
logs[i] is guaranteed to have an identifier, and a word after the identifier.
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class Solution:
def reorderLogFiles(self, logs: List[str]) -> List[str]:
nums, letters = [], []
for log in logs:
words = log.split(' ')
if words[1].isdigit():
nums.append(log)
else:
letters.append(log)
letters.sort(key = lambda x: x.split(' ')[0])
letters.sort(key = lambda x: x.split(' ')[1:])
return letters + nums

面经:Amazon,用key function,sort是stable

189. Rotate Array (Easy)

Given an array, rotate the array to the right by k steps, where k is non-negative.

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Example 1:
Input: [1,2,3,4,5,6,7] and k = 3
Output: [5,6,7,1,2,3,4]
Explanation:
rotate 1 steps to the right: [7,1,2,3,4,5,6]
rotate 2 steps to the right: [6,7,1,2,3,4,5]
rotate 3 steps to the right: [5,6,7,1,2,3,4]

Example 2:
Input: [-1,-100,3,99] and k = 2
Output: [3,99,-1,-100]
Explanation:
rotate 1 steps to the right: [99,-1,-100,3]
rotate 2 steps to the right: [3,99,-1,-100]

Note:
Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
Could you do it in-place with O(1) extra space?

O(n) space:

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class Solution:
def rotate(self, nums: List[int], k: int) -> None:
n = len(nums)
original = nums[:]
for i in range(n):
nums[(i + k) % n] = original[i]

O(1) space:

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class Solution:
def rotate(self, nums: List[int], k: int) -> None:
k = k % len(nums)
if k == 0 or len(nums) == 1:
return
nums.reverse()
self.reverse(nums, 0, k - 1)
self.reverse(nums, k, len(nums) - 1)
def reverse(self, nums, s, e):
while s < e:
nums[s], nums[e] = nums[e], nums[s]
s += 1
e -= 1

面经:维萨. 向右走2:—>–> to –>—> 或者向左走2:<–<— to <—<–

296. Best Meeting Point (Hard) 带锁

LinC 912. Best Meeting Point
A group of two or more people wants to meet and minimize the total travel distance. You are given a 2D grid of values 0 or 1, where each 1 marks the home of someone in the group. The distance is calculated using Manhattan Distance, where distance(p1, p2) = |p2.x - p1.x| + |p2.y - p1.y|.

Example 1:
Input:
[[1,0,0,0,1],[0,0,0,0,0],[0,0,1,0,0]]
Output:
6
Explanation:
The point (0,2) is an ideal meeting point, as the total travel distance of 2 + 2 + 2 = 6 is minimal. So return 6.

Example 2:
Input:
[[1,1,0,0,1],[1,0,1,0,0],[0,0,1,0,1]]
Output:
14

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class Solution:
def minTotalDistance(self, grid):
rows, cols = [], []
for r in range(len(grid)):
for c in range(len(grid[0])):
if grid[r][c] == 1:
rows.append(r)
cols.append(c)
rows.sort()
cols.sort()
ans = 0
i, j = 0, len(rows) - 1
while i < j:
ans += rows[j] - rows[i] + cols[j] - cols[i]
i += 1
j -= 1
return ans

面经:维萨。关键是知道将行和列排序,然后从两边算距离的方法。

Sliding Window 滑动窗口

438. Find All Anagrams in a String (Medium)

Given a string s and a non-empty string p, find all the start indices of p’s anagrams in s.
Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.
The order of output does not matter.

Example 1:
Input:
s: “cbaebabacd” p: “abc”
Output:
[0, 6]
Explanation:
The substring with start index = 0 is “cba”, which is an anagram of “abc”.
The substring with start index = 6 is “bac”, which is an anagram of “abc”.

Example 2:
Input:
s: “abab” p: “ab”
Output:
[0, 1, 2]
Explanation:
The substring with start index = 0 is “ab”, which is an anagram of “ab”.
The substring with start index = 1 is “ba”, which is an anagram of “ab”.
The substring with start index = 2 is “ab”, which is an anagram of “ab”.

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class Solution:
def findAnagrams(self, s: str, p: str) -> List[int]:
ans = []
n, m = len(s), len(p)
from collections import Counter
pCounter = Counter(p)
sCounter = Counter(s[:m - 1])
for i in range(m - 1, n):
si = i - m + 1
sCounter[s[i]] += 1
if pCounter == sCounter:
ans.append(si)
sCounter[s[si]] -= 1
if sCounter[s[si]] == 0:
del sCounter[s[si]]
return ans

面经:Amazon。注意sCounter[s[si]] -= 1需要减相应字母的计数,否则会产生bug。todo:也用sliding window的模板改下代码

3. Longest Substring Without Repeating Characters (Medium)

Given a string, find the length of the longest substring without repeating characters.

Example 1:
Input: “abcabcbb”
Output: 3
Explanation: The answer is “abc”, with the length of 3.

Example 2:
Input: “bbbbb”
Output: 1
Explanation: The answer is “b”, with the length of 1.

Example 3:
Input: “pwwkew”
Output: 3
Explanation: The answer is “wke”, with the length of 3.
Note that the answer must be a substring, “pwke” is a subsequence and not a substring.

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class Solution:
def lengthOfLongestSubstring(self, s: str) -> int:
l, d = 0, {}
ans = 0
for i, c in enumerate(s):
if c in d:
l = max(l, d[c] + 1)
d[c] = i
ans = max(ans, i - l + 1)
return ans

面经,二刷:Amazon。sliding window模板代码看不懂了。l和i两个指针,如果c重复了,就将l挪到上次出现的c后一位. l不能往回走,如abba,到最后a的时候l不能回到0+1的位置上去

76. Minimum Window Substring (Hard)

Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).

Example:
Input: S = “ADOBECODEBANC”, T = “ABC”
Output: “BANC”

Note:
If there is no such window in S that covers all characters in T, return the empty string “”.
If there is such window, you are guaranteed that there will always be only one unique minimum window in S.

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class Solution:
def minWindow(self, s: str, t: str) -> str:
from collections import Counter, defaultdict
targetMap = Counter(t)
window = defaultdict(int)
l, r = 0, 0
targetLen, formed = len(targetMap), 0
ans = ""
#search left to right
while r <= len(s) - 1:
c = s[r]
window[c] += 1
if c in targetMap and window[c] == targetMap[c]:
formed += 1
#found a valid substring, contracts left
while formed == targetLen:
if not ans or r - l + 1 < len(ans):
ans = s[l:r + 1]
tempC = s[l]
window[tempC] -= 1
if tempC in targetMap and window[tempC] < targetMap[tempC]:
formed -= 1
l += 1
r += 1
return ans

面经, 二刷:Amazon。双指针sliding window,r前进,找到符合条件的substring以后收缩l寻找更短符合条件的substring

1151 Minimum Swaps to Group All 1’s Together (Medium) 带锁

geeks for geeks article
Given an array of 0’s and 1’s, we need to write a program to find the minimum number of swaps required to group all 1’s present in the array together.

Examples:
Input : arr[] = {1, 0, 1, 0, 1}
Output : 1
Explanation: Only 1 swap is required to
group all 1’s together. Swapping index 1
and 4 will give arr[] = {1, 1, 1, 0, 0}

Input : arr[] = {1, 0, 1, 0, 1, 1}
Output : 1

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def minSwaps(arr):
from collections import Counter
counter = Counter(arr)
targetLen = counter[1]
counter = Counter(arr[0:targetLen])
most1s = counter[1]
for i in range(1, len(arr) - targetLen + 1):
#because I already know how many 1s in previous subarray. Just add or subtract as I go
if boxes[i - 1] == 1:
counter[1] -= 1
if boxes[i + targetLen - 1] == 1:
counter[1] += 1
if counter[1] > most1s:
most1s = counter[1]
return targetLen - most1s

面经:Celo。算法的核心是:find the subarray of counter[1] length that has the most 1’s. Then just move the 1’s from elsewhere to fill the 0’s in this subarray

1004. Max Consecutive Ones III medium

Given an array A of 0s and 1s, we may change up to K values from 0 to 1.
Return the length of the longest (contiguous) subarray that contains only 1s.

Example 1:
Input: A = [1,1,1,0,0,0,1,1,1,1,0], K = 2
Output: 6
Explanation:
[1,1,1,0,0,1,1,1,1,1,1]
Bolded numbers were flipped from 0 to 1. The longest subarray is underlined.

Example 2:
Input: A = [0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1], K = 3
Output: 10
Explanation:
[0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,1]
Bolded numbers were flipped from 0 to 1. The longest subarray is underlined.

Note:
1 <= A.length <= 20000
0 <= K <= A.length
A[i] is 0 or 1

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class Solution:
def longestOnes(self, A: List[int], K: int) -> int:
l = 0
for r in range(len(A)):
K -= 1 - A[r]
if K < 0:
K += 1 - A[l]
l += 1
return r - l + 1

一刷:Facebook tag

Linked List 链表

83. Remove Duplicates from Sorted List (Easy)

Given a sorted linked list, delete all duplicates such that each element appear only once.

Example 1:
Input: 1->1->2
Output: 1->2

Example 2:
Input: 1->1->2->3->3
Output: 1->2->3

高频

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class Solution:
def deleteDuplicates(self, head: ListNode) -> ListNode:
cur = head
while cur and cur.next:
if cur.val == cur.next.val:
cur.next = cur.next.next
else:
cur = cur.next
return head

总结:…temp = head; while temp and temp.next: …temp.next = temp.next.next; else: temp = temp.next…

82. Remove Duplicates from Sorted List II (Medium)

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.

Example 1:
Input: 1->2->3->3->4->4->5
Output: 1->2->5

Example 2:
Input: 1->1->1->2->3
Output: 2->3

高频

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class Solution:
def deleteDuplicates(self, head: ListNode) -> ListNode:
dummy = cur = ListNode(0)
dummy.next = head
while cur:
has_dupe = False
while cur.next and cur.next.next and cur.next.val == cur.next.next.val:
cur.next = cur.next.next
has_dupe = True
if has_dupe:
cur.next = cur.next.next
else:
cur = cur.next
return dummy.next

高频:…cur = dummy…has_dupe = False…cur.next = cur.next.next…

21. Merge Two Sorted Lists (Easy)

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

Example:
Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4

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class Solution:
def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
dummy = cur = ListNode(0)
while l1 and l2:
if l1.val < l2.val:
cur.next = l1
cur = cur.next
l1 = l1.next
else:
cur.next = l2
cur = cur.next
l2 = l2.next
if l1:
cur.next = l1
if l2:
cur.next = l2
return dummy.next

高频:面经:Cruise。
三刷:手熟尔

19. Remove Nth Node From End of List (Medium)

Given a linked list, remove the n-th node from the end of list and return its head.

Example:
Given linked list: 1->2->3->4->5, and n = 2.
After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:
Given n will always be valid.

Follow up:
Could you do this in one pass?

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class Solution:
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
dummy = ListNode(0)
dummy.next = head
fast = slow = dummy
while n:
n -= 1
fast = fast.next
while fast and fast.next:
fast = fast.next
slow = slow.next
slow.next = slow.next.next
return dummy.next

高频:dummy = ListNode(0)…fast = slow = dummy…slow.next = slow.next.next…

2. Add Two Numbers (Medium)

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You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

高频

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class Solution:
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
return self.n2l(self.l2n(l1) + self.l2n(l2))

def l2n(self, root):
num = ""
while root:
num = str(root.val) + num
root = root.next
return int(num)

def n2l(self, num):
num = str(num)
head = None
for d in num:
if not head:
head = ListNode(int(d))
else:
old_head = head
head = ListNode(int(d))
head.next = old_head
return head

总结:写l2n和n2l两个函数

86. Partition List (Medium)

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.
You should preserve the original relative order of the nodes in each of the two partitions.

Example:
Input: head = 1->4->3->2->5->2, x = 3
Output: 1->2->2->4->3->5

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class Solution:
def partition(self, head: ListNode, x: int) -> ListNode:
loHead = loDummy = ListNode(0)
hiHead = hiDummy = ListNode(0)
while head:
if head.val < x:
loHead.next = head
loHead = loHead.next
else:
hiHead.next = head
hiHead = hiHead.next
head = head.next
loHead.next = hiDummy.next
hiHead.next = None
return loDummy.next

高频, 五刷:两个dummy,两个worker

141. Linked List Cycle (Easy)

Given a linked list, determine if it has a cycle in it.
To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.

Example 1:
Input: head = [3,2,0,-4], pos = 1
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the second node.
Linked List Cycle example1

Example 2:
Input: head = [1,2], pos = 0
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the first node.
Linked List Cycle example2

Example 3:
Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list.
Linked List Cycle example3

Follow up:
Can you solve it without using extra space?

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class Solution(object):
def hasCycle(self, head):
if not head:
return False
s = f = head
while f.next and f.next.next:
s = s.next
f = f.next.next
if s == f:
return True
return False

二刷,面经:维萨。

142. Linked List Cycle II (Medium)

Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.
Note: Do not modify the linked list.

Example 1:
Input: head = [3,2,0,-4], pos = 1
Output: tail connects to node index 1
Explanation: There is a cycle in the linked list, where tail connects to the second node.
Linked List Cycle II example1

Example 2:
Input: head = [1,2], pos = 0
Output: tail connects to node index 0
Explanation: There is a cycle in the linked list, where tail connects to the first node.
Linked List Cycle II example2

Example 3:
Input: head = [1], pos = -1
Output: no cycle
Explanation: There is no cycle in the linked list.
Linked List Cycle II example3

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class Solution(object):
def detectCycle(self, head):
if not head or not head.next:
return
s = f = head
loop = False
while f.next and f.next.next:
s = s.next
f = f.next.next
if s == f:
loop = True
break
if not loop:
return
s = head
while s != f:
s = s.next
f = f.next
return s

面经:维萨。关键知识点是1.相遇点是慢指针在环里的 第一圈。是快指针在环里的第n圈。2. 慢指针走了x+y, 快指针走了 2(x + y) 3. 快指针这个 距离又等同于 x + y + nr 4. x = nr - y 4. 由下图可见, nr - y = z 5. z 和 x 是相等的。 6. 因此可以head与快或慢同时前进,相遇即是B点。

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    x            y
A ------ B --------+
| |
z | |
+----C----+
* 环的长度为 r
* C: 快慢指针相遇点

206. Reverse Linked List (Easy)

Reverse a singly linked list.

Example:
Input: 1->2->3->4->5->NULL
Output: 5->4->3->2->1->NULL

Follow up:
A linked list can be reversed either iteratively or recursively. Could you implement both?
遍历:

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class Solution:
def reverseList(self, head: ListNode) -> ListNode:
pre = None
cur = head
while cur:
nextHead = cur.next
cur.next = pre
pre = cur
cur = nextHead
return pre

递归:

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class Solution:
def reverseList(self, head: ListNode) -> ListNode:
if not head:
return head
if head and not head.next:
return head
savedListHead = self.reverseList(head.next)
head.next.next = head
head.next = None
return savedListHead

递归:

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class Solution:
def reverseList(self, head: ListNode) -> ListNode:
return self.helper(head, None)
def helper(self, head, pre):
if not head:
return pre
nextHead = head.next
head.next = pre
return self.helper(nextHead, head)

高频, 四刷:遍历:pre = None…nextHead = head.next…递归:增加了这个油管视频的递归方法。savedListHead是base case,然后最后层层传回去。…return f(nextHead, head)…

160. Intersection of Two Linked Lists (Easy)

Write a program to find the node at which the intersection of two singly linked lists begins.
For example, the following two linked lists:

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A:          a1 → a2

c1 → c2 → c3

B: b1 → b2 → b3

begin to intersect at node c1.

Notes:
If the two linked lists have no intersection at all, return null.
The linked lists must retain their original structure after the function returns.
You may assume there are no cycles anywhere in the entire linked structure.
Your code should preferably run in O(n) time and use only O(1) memory.

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class Solution(object):
def getIntersectionNode(self, headA, headB):
savedHeadA, savedHeadB = headA, headB
if headA == None or headB == None:
return None
lenA, lenB = 1, 1
while headA.next != None:
headA = headA.next
lenA += 1
while headB.next != None:
headB = headB.next
lenB += 1
if lenA > lenB:
diff = lenA - lenB
for x in xrange(diff):
savedHeadA = savedHeadA.next
else:
diff = lenB - lenA
for x in xrange(diff):
savedHeadB = savedHeadB.next
while savedHeadA != None:
if savedHeadA == savedHeadB:
return savedHeadA
else:
savedHeadA = savedHeadA.next
savedHeadB = savedHeadB.next
return None

思路:统计两条链走到头的长度,lenA 和 lenB, 然后让长的那条先走两者的差值,然后一起走,返回相遇的那点。或者将A的尾巴连到B的开头,找环的入口
总结:1.注意空输入(不能假设 headA 或 B 有 next)2.注意 headA headB 是一个节点 i.e. 合体的情况

24. Swap Nodes in Pairs (Medium)

Given a linked list, swap every two adjacent nodes and return its head.
You may not modify the values in the list’s nodes, only nodes itself may be changed.

Example:
Given 1->2->3->4, you should return the list as 2->1->4->3.

高频

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class Solution:
def swapPairs(self, head: ListNode) -> ListNode:
dummy = cur = ListNode(0)
while head and head.next:
a, b = head, head.next
head = b.next
cur.next = b
b.next = a
cur = a
cur.next = None
if head:
cur.next = head
return dummy.next

递归

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class Solution:
def swapPairs(self, head: ListNode) -> ListNode:
if not head or not head.next:
return head
neighbor = head.next
frontier = neighbor.next
neighbor.next = head
head.next = self.swapPairs(frontier)
return neighbor

总结:a, b, 挪head,连接 cur.next,b的next到a,挪cur到a,切掉cur->。todo:理解递归答案

61. Rotate List (Medium)

Given a linked list, rotate the list to the right by k places, where k is non-negative.

Example 1:
Input: 1->2->3->4->5->NULL, k = 2
Output: 4->5->1->2->3->NULL
Explanation:
rotate 1 steps to the right: 5->1->2->3->4->NULL
rotate 2 steps to the right: 4->5->1->2->3->NULL

Example 2:
Input: 0->1->2->NULL, k = 4
Output: 2->0->1->NULL
Explanation:
rotate 1 steps to the right: 2->0->1->NULL
rotate 2 steps to the right: 1->2->0->NULL
rotate 3 steps to the right: 0->1->2->NULL
rotate 4 steps to the right: 2->0->1->NULL

高频

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class Solution:
def rotateRight(self, head: ListNode, k: int) -> ListNode:
if not head:
return head
dummy = ListNode(0)
dummy.next = head
f = dummy
l = 0
while f and f.next:
f = f.next
l += 1
s = dummy
k %= l
for _ in range(l - k):
s = s.next
f.next = dummy.next
dummy.next = s.next
s.next = None
return dummy.next

高频,二刷:链长l,k %= l, 将第l - k位置的以后的放链表前面:s走l - k步,f.next接到dummy.next上,dummy.next接到s.next; s.next = None

146. LRU Cache (Medium)

Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and put.

get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
put(key, value) - Set or insert the value if the key is not already present. When the cache reached its capacity, it should invalidate the least recently used item before inserting a new item.
The cache is initialized with a positive capacity.

Follow up:
Could you do both operations in O(1) time complexity?

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Example:
LRUCache cache = new LRUCache( 2 /* capacity */ );
cache.put(1, 1);
cache.put(2, 2);
cache.get(1); // returns 1
cache.put(3, 3); // evicts key 2
cache.get(2); // returns -1 (not found)
cache.put(4, 4); // evicts key 1
cache.get(1); // returns -1 (not found)
cache.get(3); // returns 3
cache.get(4); // returns 4

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class Node:
def __init__(self, k, v):
self.key = k
self.val = v
self.prev = None
self.next = None

class LRUCache:

def __init__(self, capacity: int):
self.capacity = capacity
self.mapping = {}
self.head, self.tail = Node(0, 0), Node(0, 0)
self.head.next = self.tail
self.tail.prev = self.head

def get(self, key: int) -> int:
if key not in self.mapping:
return -1
n = self.mapping[key]
self._remove(n)
self._append(n)
return n.val

def put(self, key: int, value: int) -> None:
n = Node(key, value)
if key in self.mapping:
self._remove(self.mapping[key])
self.mapping[key] = n
self._append(n)
if len(self.mapping) > self.capacity:
t = self.head.next
del self.mapping[t.key]
self._remove(t)

def _remove(self, node):
node.next.prev = node.prev
node.prev.next = node.next

def _append(self, node):
self.tail.prev.next = node
node.prev = self.tail.prev
node.next = self.tail
self.tail.prev = node

高频, 二刷:dict的k为k,v为链表的节点n。注意put的时候要先把旧的重复key对从双链表中删除。或者使用OrderedDict这个数据结构,支持随机+头部删除(mapping.popitem(last = False))和顺序append

234. Palindrome Linked List (Easy)

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Given a singly linked list, determine if it is a palindrome.

Example 1:

Input: 1->2
Output: false
Example 2:

Input: 1->2->2->1
Output: true
Follow up:
Could you do it in O(n) time and O(1) space?
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class Solution:
def isPalindrome(self, head: ListNode) -> bool:
if not head or not head.next:
return True
s = f = head
midHead = None
while f and f.next:
f = f.next.next
nextHead = s.next
s.next = midHead
midHead = s
s = nextHead
if f:
s = s.next
while s:
if midHead.val != s.val:
return False
else:
s = s.next
midHead = midHead.next
return True

高频, 二刷:慢指针边走边反转,当链表是奇数个时需要s跳过中间节点再s,midhead同时前进:…if f: s = s.next…

Stack 栈

20. Valid Parentheses (Easy)

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Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.

An input string is valid if:

Open brackets must be closed by the same type of brackets.
Open brackets must be closed in the correct order.
Note that an empty string is also considered valid.

Example 1:

Input: "()"
Output: true
Example 2:

Input: "()[]{}"
Output: true
Example 3:

Input: "(]"
Output: false
Example 4:

Input: "([)]"
Output: false
Example 5:

Input: "{[]}"
Output: true
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class Solution:
def isValid(self, s: str) -> bool:
l = '({['
stack = []
for c in s:
if c in l:
stack.append(c)
else:
if not stack:
return False
t = stack.pop()
if t == '(' and c != ')':
return False
if t == '{' and c != '}':
return False
if t == '[' and c != ']':
return False
if stack:
return False
return True

高频:注意在出栈前检验栈是否为空,走完以后检查栈是否为空

71. Simplify Path (Medium)

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Given an absolute path for a file (Unix-style), simplify it. Or in other words, convert it to the canonical path.

In a UNIX-style file system, a period . refers to the current directory. Furthermore, a double period .. moves the directory up a level. For more information, see: Absolute path vs relative path in Linux/Unix

Note that the returned canonical path must always begin with a slash /, and there must be only a single slash / between two directory names. The last directory name (if it exists) must not end with a trailing /. Also, the canonical path must be the shortest string representing the absolute path.

Example 1:

Input: "/home/"
Output: "/home"
Explanation: Note that there is no trailing slash after the last directory name.
Example 2:

Input: "/../"
Output: "/"
Explanation: Going one level up from the root directory is a no-op, as the root level is the highest level you can go.
Example 3:

Input: "/home//foo/"
Output: "/home/foo"
Explanation: In the canonical path, multiple consecutive slashes are replaced by a single one.
Example 4:

Input: "/a/./b/../../c/"
Output: "/c"
Example 5:

Input: "/a/../../b/../c//.//"
Output: "/c"
Example 6:

Input: "/a//b////c/d//././/.."
Output: "/a/b/c"
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class Solution:
def simplifyPath(self, path: str) -> str:
t = path.split('/')
s = []
for p in t:
if p and p != '.':
if p == '..':
if s:
s.pop()
else:
s.append(p)
if not s:
return '/'
else:
return '/' + '/'.join(s)

高频:注意split(‘/‘)是’/‘的会变成数组里一个空的元素

150. Evaluate Reverse Polish Notation (Medium)

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Evaluate the value of an arithmetic expression in Reverse Polish Notation.

Valid operators are +, -, *, /. Each operand may be an integer or another expression.

Note:

Division between two integers should truncate toward zero.
The given RPN expression is always valid. That means the expression would always evaluate to a result and there won't be any divide by zero operation.
Example 1:

Input: ["2", "1", "+", "3", "*"]
Output: 9
Explanation: ((2 + 1) * 3) = 9
Example 2:

Input: ["4", "13", "5", "/", "+"]
Output: 6
Explanation: (4 + (13 / 5)) = 6
Example 3:

Input: ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"]
Output: 22
Explanation:
((10 * (6 / ((9 + 3) * -11))) + 17) + 5
= ((10 * (6 / (12 * -11))) + 17) + 5
= ((10 * (6 / -132)) + 17) + 5
= ((10 * 0) + 17) + 5
= (0 + 17) + 5
= 17 + 5
= 22
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class Solution:
def evalRPN(self, tokens: List[str]) -> int:
operators = "+-*/"
s = []
for op in tokens:
if op not in operators:
s.append(int(op))
else:
b = s.pop()
a = s.pop()
if op == "+":
res = a + b
elif op == "-":
res = a - b
elif op == '*':
res = a * b
elif op == "/":
if a * b < 0 and a % b != 0:
res = a // b + 1
else:
res = a // b
s.append(res)
return s[0]

高频:需要注意leetcode里-1//20为0,而python里为-1,-21//20python里为-2。需要单独处理一下

155. Min Stack (Easy)

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Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

push(x) -- Push element x onto stack.
pop() -- Removes the element on top of the stack.
top() -- Get the top element.
getMin() -- Retrieve the minimum element in the stack.


Example:

MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); --> Returns -3.
minStack.pop();
minStack.top(); --> Returns 0.
minStack.getMin(); --> Returns -2.
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class MinStack:

def __init__(self):
"""
initialize your data structure here.
"""
self.stack = []

def push(self, x: int) -> None:
curMin = self.getMin() if self.stack else sys.maxsize
if x < curMin:
curMin = x
self.stack.append((x, curMin))

def pop(self) -> None:
self.stack.pop()

def top(self) -> int:
return self.stack[-1][0]

def getMin(self) -> int:
if self.stack:
return self.stack[-1][1]

高频:一个stack就用tuple,两个stack就相当于第二个stack来维持curMin

224. Basic Calculator (Hard)

Implement a basic calculator to evaluate a simple expression string.
The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and empty spaces .

Example 1:
Input: “1 + 1”
Output: 2
Example 2:

Input: “ 2-1 + 2 “
Output: 3
Example 3:

Input: “(1+(4+5+2)-3)+(6+8)”
Output: 23
Note:
You may assume that the given expression is always valid.
Do not use the eval built-in library function.

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class Solution:
def calculate(self, s: str) -> int:
num, ans = 0, 0
sign = 1
stack = []
for c in s:
if c.isdigit():
num = num * 10 + int(c)
elif c in "+-":
ans += sign * num
num = 0
sign = 1 if c == "+" else -1
elif c == "(":
stack.append(ans)
stack.append(sign)
ans = 0
sign = 1
elif c == ")":
ans += sign * num
ans *= stack.pop()
ans += stack.pop()
num = 0
return ans + sign * num

面经:Cruise …num = 0…ans = 0; sign = 1…num = 0…

227. Basic Calculator II (Medium)

Implement a basic calculator to evaluate a simple expression string.
The expression string contains only non-negative integers, +, -, * , / operators and empty spaces . The integer division should truncate toward zero.

Example 1:
Input: “3+2*2”
Output: 7
Example 2:

Input: “ 3/2 “
Output: 1
Example 3:

Input: “ 3+5 / 2 “
Output: 5

Note:
You may assume that the given expression is always valid.
Do not use the eval built-in library function.

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class Solution:
def calculate(self, s: str) -> int:
if not s:
return 0
num = 0
operator = "+"
stack = []
for i in range(len(s)):
if s[i].isdigit():
num = num * 10 + int(s[i])
if (not s[i].isdigit() and s[i] != " ") or i == len(s) - 1:
if operator == "+":
stack.append(num)
elif operator == "-":
stack.append(-num)
elif operator == "*":
stack.append(stack.pop() * num)
elif operator == "/":
stack.append(int(stack.pop() / num))
num = 0
operator = s[i]
return sum(stack)

面经:Cruise。符号或操作符均是在最后变化。因为要处理最后一个字符的问题,所以要么两个if,要么得在s后面跟一个“+”

Queue 队列

LinC 642. Moving Average from Data Stream (Easy)

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Given a stream of integers and a window size, calculate the moving average of all integers in the sliding window.

Example
MovingAverage m = new MovingAverage(3);
m.next(1) = 1 // return 1.00000
m.next(10) = (1 + 10) / 2 // return 5.50000
m.next(3) = (1 + 10 + 3) / 3 // return 4.66667
m.next(5) = (10 + 3 + 5) / 3 // return 6.00000

思路:建个 window size 的队列,返回队列的平均值

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class MovingAverage:
q = collections.deque()
sum = 0
maxLen = 0
"""
@param: size: An integer
"""
def __init__(self, size):
# do intialization if necessary
self.maxLen = size
"""
@param: val: An integer
@return:
"""
def next(self, val):
# write your code here
self.q.append(val)
self.sum += val
if len(self.q) > self.maxLen:
temp = self.q.popleft()
self.sum -= temp
avg = self.sum / len(self.q)
return avg

# Your MovingAverage object will be instantiated and called as such:
# obj = MovingAverage(size)
# param = obj.next(val)

总结:注意 class 变量要加 self,另外 sum 不要每次都 loop 一遍算, 直接放到 class 变量里,每次只增 and / or 减一次。

Hash 哈希表

290. Word Pattern (Easy)

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Given a pattern and a string str, find if str follows the same pattern.

Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in str.

Example 1:

Input: pattern = "abba", str = "dog cat cat dog"
Output: true
Example 2:

Input:pattern = "abba", str = "dog cat cat fish"
Output: false
Example 3:

Input: pattern = "aaaa", str = "dog cat cat dog"
Output: false
Example 4:

Input: pattern = "abba", str = "dog dog dog dog"
Output: false
Notes:
You may assume pattern contains only lowercase letters, and str contains lowercase letters separated by a single space.

思路:关键在于懂得建立 pattern 里每个字母和 str 里每个 word 的映射。

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class Solution(object):
def wordPattern(self, pattern, str):
"""
:type pattern: str
:type str: str
:rtype: bool
"""
words = str.split(' ')
if len(words) != len(pattern):
return False
mapping = {}
for i, char in enumerate(pattern):
if char in mapping:
if mapping[char] != words[i]:
return False
else:
if words[i] in mapping.values():
return False
mapping[char] = words[i]
return True

总结:注意需要用 enumerate, 因为要同时遍历 pattern 和 str. 很好的哈希表热身题。Word Pattern II 的 str 里没有空格了,不能直接 split,难度直接推到 Hard。目前刷题的水平先跳过吧 :(

387. First Unique Character in a String (Easy)

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Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1.

Examples:

s = "leetcode"
return 0.

s = "loveleetcode",
return 2.
Note: You may assume the string contain only lowercase letters.

思路:过两遍,第一遍数出现多少次, 第二遍把第一个为 1 的 index 返回

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class Solution(object):
def firstUniqChar(self, s):
"""
:type s: str
:rtype: int
"""
if len(s) == 0:
return -1
if len(s) == 1:
return 0
dict = {}
for char in s:
if char not in dict:
dict[char] = 1
else:
dict[char] += 1
for index, char in enumerate(s):
if dict[char] == 1:
return index
return -1

总结:基本题,注意 dict entry 初始化为 1 的情况

409. Longest Palindrome (Easy)

Given a string which consists of lowercase or uppercase letters, find the length of the longest palindromes that can be built with those letters.
This is case sensitive, for example “Aa” is not considered a palindrome here.

Note:
Assume the length of given string will not exceed 1,010.

Example:
Input:
“abccccdd”
Output:
7

Explanation:
One longest palindrome that can be built is “dccaccd”, whose length is 7.

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class Solution(object):
def longestPalindrome(self, s):
"""
:type s: str
:rtype: int
"""
ans = 0
from collections import Counter
counter = Counter(s)
for c, v in counter.items():
if v % 2 == 0 or ans % 2 == 0:
ans += v
else:
ans += v - 1
return ans

面经:Amazon。去掉了一刷二刷的hashmap和dp解法。就用简单好懂的解法吧

380. Insert Delete GetRandom O(1) (Medium)

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Design a data structure that supports all following operations in average O(1) time.

insert(val): Inserts an item val to the set if not already present.
remove(val): Removes an item val from the set if present.
getRandom: Returns a random element from current set of elements. Each element must have the same probability of being returned.
Example:

// Init an empty set.
RandomizedSet randomSet = new RandomizedSet();

// Inserts 1 to the set. Returns true as 1 was inserted successfully.
randomSet.insert(1);

// Returns false as 2 does not exist in the set.
randomSet.remove(2);

// Inserts 2 to the set, returns true. Set now contains [1,2].
randomSet.insert(2);

// getRandom should return either 1 or 2 randomly.
randomSet.getRandom();

// Removes 1 from the set, returns true. Set now contains [2].
randomSet.remove(1);

// 2 was already in the set, so return false.
randomSet.insert(2);

// Since 2 is the only number in the set, getRandom always return 2.
randomSet.getRandom();

思路:看答案知道需要用 list 和 dictionary,因为要满足 O(1), 因为仅有 list 的 in 操作不能满足 O(1)

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class RandomizedSet(object):

def __init__(self):
"""
Initialize your data structure here.
"""
self.list = []
self.dict = {}

def insert(self, val):
"""
Inserts a value to the set. Returns true if the set did not already contain the specified element.
:type val: int
:rtype: bool
"""
if val in self.dict:
return False
else:
self.list.append(val)
self.dict[val] = len(self.list) - 1
return True

def remove(self, val):
"""
Removes a value from the set. Returns true if the set contained the specified element.
:type val: int
:rtype: bool
"""
if val in self.dict:
index, lastVal = self.dict[val], self.list[len(self.list) - 1]
self.list[index], self.dict[lastVal] = lastVal, index
self.list.pop()
self.dict.pop(val)
return True
else:
return False

def getRandom(self):
"""
Get a random element from the set.
:rtype: int
"""
return self.list[random.randint(0, len(self.list) - 1)]


# Your RandomizedSet object will be instantiated and called as such:
# obj = RandomizedSet()
# param_1 = obj.insert(val)
# param_2 = obj.remove(val)
# param_3 = obj.getRandom()

总结:可能是用 python 的原因,搞明白问什么了一次过

LinC 960. First Unique Number in a Stream II (Medium)

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Description
We need to implement a data structure named DataStream. There are two methods required to be implemented:

void add(number) // add a new number
int firstUnique() // return first unique number
You can assume that there must be at least one unique number in the stream when calling the firstUnique.

Example
add(1)
add(2)
firstUnique() => 1
add(1)
firstUnique() => 2

思路:维持一个 deque / queue,碰到相同的就 popleft 出去

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class DataStream:

def __init__():
# do intialization if necessary
self.q = collections.deque()
self.dict = {}
"""
@param num: next number in stream
@return: nothing
"""
def add(self, num):
# write your code here
if num in self.dict:
self.dict[num] += 1
else:
self.dict[num] = 1
self.q.append(num)
"""
@return: the first unique number in stream
"""
def firstUnique(self):
# write your code here
while len(self.q) > 0 and self.dict[self.q[0]] > 1:
self.q.popleft()
return self.q[0]

总结:1.popleft 要在 firstUnique 里面,不然有些 testcase 过不了;2.注意 popleft 的条件要用 while, 用 for 会出错

49. Group Anagrams (Medium)

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Given an array of strings, group anagrams together.

Example:

Input: ["eat", "tea", "tan", "ate", "nat", "bat"],
Output:
[
["ate","eat","tea"],
["nat","tan"],
["bat"]
]
Note:

All inputs will be in lowercase.
The order of your output does not matter.

高频

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class Solution:
def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
d = collections.defaultdict(list)
for s in strs:
tmps = "".join(sorted(s))
d[tmps].append(s)
return list(d.values())

总结:sorted(s)返回一个char list,“”.join(list)将这个list拼回字符串。list(d.values())可返回defaultdict的值

Heap (Priority Queue)

264. Ugly Number II (Medium)

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Write a program to find the n-th ugly number.

Ugly numbers are positive numbers whose prime factors only include 2, 3, 5.

Example:

Input: n = 10
Output: 12
Explanation: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 is the sequence of the first 10 ugly numbers.
Note:

1 is typically treated as an ugly number.
n does not exceed 1690.

思路:九章的 python 答案可以 work,但是实在是不好理解。写个好理解一点的版本。heapq 和 hashMap, 从 heapq 中取 n - 1 次(第一个数为 1),每取一次将原始 ugly numbers 2, 3, 5 过一遍

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class Solution(object):
def nthUglyNumber(self, n):
"""
:type n: int
:rtype: int
"""
primes = [2, 3, 5]
q = [2, 3, 5]
hashMap = {}
for i in range(3):
hashMap[q[i]] = True
import heapq
heapq.heapify(q)
ans = 1
for i in range(n - 1):
ans = heapq.heappop(q)
for j in range(3):
new_val = ans * primes[j]
if new_val not in hashMap:
heapq.heappush(q, new_val)
hashMap[new_val] = True
return ans

总结:可以 AC,也可以理解,good enough
二刷:其实是道dp题,用set和heapq也能过,但是时间上没有优势

973. K Closest Points to Origin (Medium)

We have a list of points on the plane. Find the K closest points to the origin (0, 0).
(Here, the distance between two points on a plane is the Euclidean distance.)
You may return the answer in any order. The answer is guaranteed to be unique (except for the order that it is in.)

Example 1:
Input: points = [[1,3],[-2,2]], K = 1
Output: [[-2,2]]
Explanation:
The distance between (1, 3) and the origin is sqrt(10).
The distance between (-2, 2) and the origin is sqrt(8).
Since sqrt(8) < sqrt(10), (-2, 2) is closer to the origin.
We only want the closest K = 1 points from the origin, so the answer is just [[-2,2]].

Example 2:
Input: points = [[3,3],[5,-1],[-2,4]], K = 2
Output: [[3,3],[-2,4]]
(The answer [[-2,4],[3,3]] would also be accepted.)

Note:
1 <= K <= points.length <= 10000
-10000 < points[i][0] < 10000
-10000 < points[i][1] < 10000

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class Solution:
def kClosest(self, points: List[List[int]], K: int) -> List[List[int]]:
arr = [[p[0] ** 2 + p[1] ** 2, p] for p in points]
arr.sort(key = lambda x: x[0])
return [i[1] for i in arr[:K]]

面经:Amazon。删掉一刷的思路和总结。简化代码,非常欣慰

LinC 545. Top k Largest Numbers II (Medium)

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Implement a data structure, provide two interfaces:
add(number). Add a new number in the data structure.
topk(). Return the top k largest numbers in this data structure. k is given when we create the data structure.

Example
s = new Solution(3);
>> create a new data structure.
s.add(3)
s.add(10)
s.topk()
>> return [10, 3]
s.add(1000)
s.add(-99)
s.topk()
>> return [1000, 10, 3]
s.add(4)
s.topk()
>> return [1000, 10, 4]
s.add(100)
s.topk()
>> return [1000, 100, 10]

思路:看着是非常直观的 min heap 问题。。。

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class Solution:
"""
@param: k: An integer
"""
def __init__(self, k):
# do intialization if necessary
self.k = k
self.q = []
"""
@param: num: Number to be added
@return: nothing
"""
def add(self, num):
# write your code here
import heapq
heapq.heappush(self.q, num)
if len(self.q) > self.k:
heapq.heappop(self.q)
"""
@return: Top k element
"""
def topk(self):
# write your code here
return sorted(self.q, reverse = True)

总结:一句 sorted(self.q, reverse = True) 完爆。。。哎, python 的 buit-in function 返回一个 sorted new list…学习了。

LinC 486. Merge K Sorted Arrays (Medium)

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Given k sorted integer arrays, merge them into one sorted array.

Example
Given 3 sorted arrays:

[
[1, 3, 5, 7],
[2, 4, 6],
[0, 8, 9, 10, 11]
]
return [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11].

Challenge
Do it in O(N log k).

N is the total number of integers.
k is the number of arrays.

思路:看答案,用 heap 屌爆了。加了链接到上面 heap 的部分

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class Solution:
"""
@param arrays: k sorted integer arrays
@return: a sorted array
"""
def mergekSortedArrays(self, arrays):
# write your code here
import heapq
q = []
for level, array in enumerate(arrays):
if len(array) == 0:
continue
heapq.heappush(q, (array[0], level, 0))
ans = []
while q:
cur, level, index = heapq.heappop(q)
ans.append(cur)
if index + 1 < len(arrays[level]):
heapq.heappush(q, (arrays[level][index + 1], level, index + 1))
return ans

总结:只能说 python 的 heapq 屌爆了

23. Merge k Sorted Lists (Hard)

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

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Example:
Input:
[
1->4->5,
1->3->4,
2->6
]
Output: 1->1->2->3->4->4->5->6

PriorityQueue:

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class Solution:
def mergeKLists(self, lists: List[ListNode]) -> ListNode:
dummy = cur = ListNode(0)
h = []
import heapq
for i, n in enumerate(lists):
if n:
heapq.heappush(h, (n.val, i, n))
while h:
v, i, n = heapq.heappop(h)
cur.next = n
cur = cur.next
if n.next:
heapq.heappush(h, (n.next.val, i, n.next))
return dummy.next

分治:

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class Solution:
def mergeKLists(self, lists: List[ListNode]) -> ListNode:
if not lists:
return None
n = len(lists)
if n == 1:
return lists[0]
mid = n // 2
l = self.mergeKLists(lists[:mid])
r = self.mergeKLists(lists[mid:])
def merge(l, r):
dummy = cur = ListNode(0)
while l and r:
if l.val < r.val:
cur.next = l
l = l.next
else:
cur.next = r
r = r.next
cur = cur.next
cur.next = l if not r else r
return dummy.next
return merge(l, r)

三种方法,都需要练习. 方法一:使用 PriorityQueue 方法二:类似归并排序的分治算法 方法三:自底向上的两两归并算法. 时间复杂度均为 O(NlogK) Strong Hire: 能够用至少2种方法进行实现,代码无大 BUG
高频:如果不能加lt(),就用(n.val, i, n)tuple防v.val重复的。
todo 把缺的第三种方法补了

295. Find Median from Data Stream (Hard)

Median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value. So the median is the mean of the two middle value.

For example,
[2,3,4], the median is 3
[2,3], the median is (2 + 3) / 2 = 2.5

Design a data structure that supports the following two operations:
void addNum(int num) - Add a integer number from the data stream to the data structure.
double findMedian() - Return the median of all elements so far.

Example:
addNum(1)
addNum(2)
findMedian() -> 1.5
addNum(3)
findMedian() -> 2

Follow up:
If all integer numbers from the stream are between 0 and 100, how would you optimize it?
If 99% of all integer numbers from the stream are between 0 and 100, how would you optimize it?

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class MedianFinder:
import heapq
def __init__(self):
self.upper = []
self.lower = []

def addNum(self, num: int) -> None:
if len(self.upper) == len(self.lower):
heapq.heappush(self.upper, -heapq.heappushpop(self.lower, -num))
else:
heapq.heappush(self.lower, -heapq.heappushpop(self.upper, num))

def findMedian(self) -> float:
if len(self.upper) == len(self.lower):
return self.upper[0] * 0.5 - self.lower[0] * 0.5
else:
return float(self.upper[0])

面经:Amazon,maxheap + minheap,lower要用负数来模拟maxheap。 注意:…heapq.heappushpop(lower, -num)…

743. Network Delay Time (Medium)

There are N network nodes, labelled 1 to N.
Given times, a list of travel times as directed edges times[i] = (u, v, w), where u is the source node, v is the target node, and w is the time it takes for a signal to travel from source to target.
Now, we send a signal from a certain node K. How long will it take for all nodes to receive the signal? If it is impossible, return -1.
network example
Example 1:
Input: times = [[2,1,1],[2,3,1],[3,4,1]], N = 4, K = 2
Output: 2
Note:
N will be in the range [1, 100].
K will be in the range [1, N].
The length of times will be in the range [1, 6000].
All edges times[i] = (u, v, w) will have 1 <= u, v <= N and 0 <= w <= 100.

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class Solution:
def networkDelayTime(self, times: List[List[int]], N: int, K: int) -> int:
graph = collections.defaultdict(dict)
for u, v, w in times:
graph[u - 1][v - 1] = w
distances = [sys.maxsize] * N
distances[K - 1] = 0
pq = [(0, K - 1)]
seen = set()
while pq:
dist, v1 = heapq.heappop(pq)
if v1 not in seen:
seen.add(v1)
for v2 in graph[v1]:
if v2 not in seen:
prev = distances[v2]
cur = dist + graph[v1][v2]
if cur < prev:
distances[v2] = cur
heapq.heappush(pq, (cur, v2))
ans = max(distances)
return -1 if ans == sys.maxsize else ans

面经:练习某家小公司的OA,Dijkstra Graph 最短路径,Dijkstra有很多变种问题,不同写法。

787. Cheapest Flights Within K Stops (Medium)

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There are n cities connected by m flights. Each fight starts from city u and arrives at v with a price w.

Now given all the cities and flights, together with starting city src and the destination dst, your task is to find the cheapest price from src to dst with up to k stops. If there is no such route, output -1.

Example 1:
Input:
n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]]
src = 0, dst = 2, k = 1
Output: 200
Explanation:
The graph looks like this:

The cheapest price from city 0 to city 2 with at most 1 stop costs 200, as marked red in the picture.
Example 2:
Input:
n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]]
src = 0, dst = 2, k = 0
Output: 500
Explanation:
The graph looks like this:

The cheapest price from city 0 to city 2 with at most 0 stop costs 500, as marked blue in the picture.
Note:

The number of nodes n will be in range [1, 100], with nodes labeled from 0 to n - 1.
The size of flights will be in range [0, n * (n - 1) / 2].
The format of each flight will be (src, dst, price).
The price of each flight will be in the range [1, 10000].
k is in the range of [0, n - 1].
There will not be any duplicated flights or self cycles.
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class Solution:
def findCheapestPrice(self, n: int, flights: List[List[int]], src: int, dst: int, K: int) -> int:
graph = collections.defaultdict(dict)
for u, v, w in flights:
graph[u][v] = w
costs = [sys.maxsize] * n
costs[src] = 0
pq = [(0, src, 0)]
while pq:
import heapq
cost, v1, stop = heapq.heappop(pq)
if stop - 1 <= K:
if v1 == dst:
return cost
for v2 in graph[v1]:
pre = costs[v2]
cur = cost + graph[v1][v2]
if cur < pre:
heapq.heappush(pq, (cur, v2, stop + 1))
return -1

面经:与上题不同之处是这里问的不是全局最优,而是符合条件的最优,同一个点可能需要重复访问才能找到符合条件的最优,因此无需seen这个set

855. Exam Room (Medium)

In an exam room, there are N seats in a single row, numbered 0, 1, 2, …, N-1.
When a student enters the room, they must sit in the seat that maximizes the distance to the closest person. If there are multiple such seats, they sit in the seat with the lowest number. (Also, if no one is in the room, then the student sits at seat number 0.)
Return a class ExamRoom(int N) that exposes two functions: ExamRoom.seat() returning an int representing what seat the student sat in, and ExamRoom.leave(int p) representing that the student in seat number p now leaves the room. It is guaranteed that any calls to ExamRoom.leave(p) have a student sitting in seat p.

Example 1:
Input: [“ExamRoom”,”seat”,”seat”,”seat”,”seat”,”leave”,”seat”], [[10],[],[],[],[],[4],[]]
Output: [null,0,9,4,2,null,5]
Explanation:
ExamRoom(10) -> null
seat() -> 0, no one is in the room, then the student sits at seat number 0.
seat() -> 9, the student sits at the last seat number 9.
seat() -> 4, the student sits at the last seat number 4.
seat() -> 2, the student sits at the last seat number 2.
leave(4) -> null
seat() -> 5, the student sits at the last seat number 5.

Note:
1 <= N <= 10^9
ExamRoom.seat() and ExamRoom.leave() will be called at most 10^4 times across all test cases.
Calls to ExamRoom.leave(p) are guaranteed to have a student currently sitting in seat number p.

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class ExamRoom:
import heapq
def __init__(self, N: int):
self.N = N
self.L = []

def seat(self) -> int:
if not self.L:
res = 0
else:
d, res = self.L[0], 0
for x, y in zip(self.L, self.L[1:]):
if (y - x) // 2 > d:
d = (y - x) // 2
res = x + (y - x) // 2
if self.N - 1 - self.L[-1] > d:
res = self.N - 1
bisect.insort(self.L, res)
return res

def leave(self, p: int) -> None:
self.L.remove(p)

面经: Cruise。正常思路是得用PriorityQueue。但是corner case写不出来,讨论区答案里java用个特殊的数据结构搭配pq,加上一个trick,不适合模板解题。…if y - x // 2 > d:…if self.N - 1 - self.L[-1] > d: res = self.N - 1…bisect.insort(…)…

Trie

208. Implement Trie (Prefix Tree) (Medium)

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Implement a trie with insert, search, and startsWith methods.

Example:

Trie trie = new Trie();

trie.insert("apple");
trie.search("apple"); // returns true
trie.search("app"); // returns false
trie.startsWith("app"); // returns true
trie.insert("app");
trie.search("app"); // returns true
Note:

You may assume that all inputs are consist of lowercase letters a-z.
All inputs are guaranteed to be non-empty strings.

思路:没啥思路,看答案

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class TrieNode:
# Initialize your data structure here.
def __init__(self):
self.word=False
self.children={}

class Trie:

def __init__(self):
self.root = TrieNode()

# @param {string} word
# @return {void}
# Inserts a word into the trie.
def insert(self, word):
node=self.root
for i in word:
if i not in node.children:
node.children[i]=TrieNode()
node=node.children[i]
node.word=True

# @param {string} word
# @return {boolean}
# Returns if the word is in the trie.
def search(self, word):
node=self.root
for i in word:
if i not in node.children:
return False
node=node.children[i]
return node.word

# @param {string} prefix
# @return {boolean}
# Returns if there is any word in the trie
# that starts with the given prefix.
def startsWith(self, prefix):
node=self.root
for i in prefix:
if i not in node.children:
return False
node=node.children[i]
return True

# Your Trie object will be instantiated and called as such:
# obj = Trie()
# obj.insert(word)
# param_2 = obj.search(word)
# param_3 = obj.startsWith(prefix)

总结:1.需要加 TrieNode 2. Leetcode 的 python3 找不到 class 是个 bug

211. Add and Search Word - Data structure design (Medium)

Design a data structure that supports the following two operations:

void addWord(word)
bool search(word)
search(word) can search a literal word or a regular expression string containing only letters a-z or .. A . means it can represent any one letter.

Example:

addWord(“bad”)
addWord(“dad”)
addWord(“mad”)
search(“pad”) -> false
search(“bad”) -> true
search(“.ad”) -> true
search(“b..”) -> true
Note:
You may assume that all words are consist of lowercase letters a-z.

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class TrieNode:
def __init__(self):
self.end = False
self.children = {}

class WordDictionary:
def __init__(self):
"""
Initialize your data structure here.
"""
self.root = TrieNode()

def addWord(self, word: str) -> None:
"""
Adds a word into the data structure.
"""
root = self.root
for c in word:
if c not in root.children:
root.children[c] = TrieNode()
root = root.children[c]
root.end = True

def search(self, word: str) -> bool:
"""
Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter.
"""
self.ans = False
self.helper(self.root, word)
return self.ans
def helper(self, root, word):
if not word:
if root.end:
self.ans = True
return
if word[0] == ".":
for v in root.children.values():
self.helper(v, word[1:])
else:
if word[0] not in root.children:
return
self.helper(root.children[word[0]], word[1:])

面经:Celo。insert:…root.end = True…

Union Find / MST

LinC 629. Minimum Spanning Tree (Hard)

Given a list of Connections, which is the Connection class (the city name at both ends of the edge and a cost between them), find edges that can connect all the cities and spend the least amount.
Return the connects if can connect all the cities, otherwise return empty list.

Example
Example 1:

Input:
[“Acity”,”Bcity”,1]
[“Acity”,”Ccity”,2]
[“Bcity”,”Ccity”,3]
Output:
[“Acity”,”Bcity”,1]
[“Acity”,”Ccity”,2]
Example 2:

Input:
[“Acity”,”Bcity”,2]
[“Bcity”,”Dcity”,5]
[“Acity”,”Dcity”,4]
[“Ccity”,”Ecity”,1]
Output:
[]

Explanation:
No way
Notice
Return the connections sorted by the cost, or sorted city1 name if their cost is same, or sorted city2 if their city1 name is also same.

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'''
Definition for a Connection
class Connection:

def __init__(self, city1, city2, cost):
self.city1, self.city2, self.cost = city1, city2, cost
'''
class Union:
def __init__(self, n):
self.size = n
self.graph = {}
for i in range(n):
self.graph[i] = i

def query(self, v1, v2):
return self.find(v1) == self.find(v2)

def find(self, v):
if self.graph[v] == v:
return v
self.graph[v] = self.find(self.graph[v])
return self.graph[v]

def connect(self, v1, v2):
root_a = self.find(v1)
root_b = self.find(v2)
if root_a != root_b:
self.size -= 1
self.graph[root_a] = self.graph[root_b]

def all_connected(self):
return self.size == 1

class Solution:
# @param {Connection[]} connections given a list of connections
# include two cities and cost
# @return {Connection[]} a list of connections from results
def lowestCost(self, connections):
# Write your code here
connections.sort(key = lambda x: x.city2)
connections.sort(key = lambda x: x.city1)
connections.sort(key = lambda x: x.cost)
citymap = {}
cnt = 0
ans = []
for c in connections:
if c.city1 not in citymap:
citymap[c.city1] = cnt
cnt += 1
if c.city2 not in citymap:
citymap[c.city2] = cnt
cnt += 1
union = Union(cnt)
for c in connections:
c1 = citymap[c.city1]
c2 = citymap[c.city2]

if not union.query(c1, c2):
union.connect(c1, c2)
ans.append(c)
return ans if union.all_connected() else []

面经:Amazon。

DP Dynamic Programming 动态规划

70. Climbing Stairs (Easy)

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You are climbing a stair case. It takes n steps to reach to the top.

Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?

Note: Given n will be a positive integer.

Example 1:

Input: 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps
Example 2:

Input: 3
Output: 3
Explanation: There are three ways to climb to the top.
1. 1 step + 1 step + 1 step
2. 1 step + 2 steps
3. 2 steps + 1 step

思路:f[n] 是为 n 时的方案数,f[1] 是为 1 时的方案数 = 1。那么爬到第n层的方法要么是从第n-1层一步上来的,要不就是从n-2层2步上来的,所以递推公式非常容易的就得出了:f[n] = f[n - 1] + f[n - 2]

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class Solution:
def climbStairs(self, n):
"""
:type n: int
:rtype: int
"""
if n <= 2:
return n
f = [0] * (n + 1)
f[1] = 1
f[2] = 2
for i in range(3, n + 1):
f[i] = f[i - 1] + f[i - 2]
return f[n]

总结:和三年前比没有变化,呵呵呵
高频:没有太多优化空间,背:爬到第n层的方法要么是从第n-1层1步上来的,要不就是从n-2层2步上来的

120. Triangle (Medium)

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Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.

For example, given the following triangle

[
[2],
[3,4],
[6,5,7],
[4,1,8,3]
]
The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11).

Note:

Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle.

思路:看了 top down 的 DP, 还是比较好理解的。f 代表到达 row 和 col 位置的最小 sum,f[i][j] 和 f[i - 1][j - 1] 的关系是:f[i][j] = mins(f[i - 1][j - 1], f[i - 1][j]) + triangle[i][j]. 规划的目标是最后一行中的最小值。;DP 以外还有三种解法,DFS:Traverse, DFS:Divide and Conquer, DFS:Divide and Conquer 加 memorization

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class Solution:
def minimumTotal(self, triangle):
"""
:type triangle: List[List[int]]
:rtype: int
"""
if len(triangle) == 0:
return 0
if len(triangle) == 1:
return triangle[0][0]
f = []
f.append([triangle[0][0]])
n = len(triangle)
for i in range(1, n):
f.append([0] * (i + 1))
for i in range(1, n):
f[i][0] = f[i - 1][0] + triangle[i][0]
f[i][i] = f[i - 1][i - 1] + triangle[i][i]
for row in range(2, n):
for col in range(1, row):
f[row][col] = min(f[row - 1][col - 1], f[row - 1][col]) + triangle[row][col]
ans = f[n - 1][0]
for i in range(1, n):
ans = min(ans, f[n - 1][i])
return ans

高频:

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class Solution:
def minimumTotal(self, triangle: List[List[int]]) -> int:
n = len(triangle)
f = [[0] * (i + 1) for i in range(n)]
f[0][0] = triangle[0][0]
for i in range(1, n):
for j in range(i + 1):
if j == 0:
f[i][0] = f[i - 1][0] + triangle[i][j]
elif j == i:
f[i][j] = f[i - 1][j - 1] + triangle[i][j]
else:
f[i][j] = min(f[i - 1][j - 1], f[i - 1][j]) + triangle[i][j]
return min(f[n - 1])

bottom up:

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class Solution:
def minimumTotal(self, triangle: List[List[int]]) -> int:
n = len(triangle)
if n == 1:
return triangle[0][0]
f = [[0] * (i + 1) for i in range(n)]
f[n - 1] = triangle[-1][:]
for r in range(n - 2, -1, -1):
for c in range(r + 1):
f[r][c] = min(f[r + 1][c], f[r + 1][c + 1]) + triangle[r][c]
return f[0][0]

总结:填充 f 每行第一个和最后一个的时候别忘了 + triangle[i][0] 和 triangle[i][i]
高频:注意topdown:下一层f[i][j]的时候要根据j的情况来分类判断f的值如何获得, 一刷的时候提前把三角的两条边先初始化了,循环的时候不循环那些元素。也是个好办法。代码稍微长一点。bottomup:代码简单很多。还可以进一步将空间降为O(n),因为之前算出来的f[r + 1][]都没用

121. Best Time to Buy and Sell Stock (Easy)

Say you have an array for which the ith element is the price of a given stock on day i.
If you were only permitted to complete at most one transaction (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.
Note that you cannot sell a stock before you buy one.

Example 1:
Input: [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
Not 7-1 = 6, as selling price needs to be larger than buying price.

Example 2:
Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

DP:

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class Solution:
def maxProfit(self, prices):
if not prices:
return 0
n = len(prices)
dp = [0] * n
low = prices[0]
for i in range(1, n):
if prices[i] <= prices[i - 1]:
dp[i] = dp[i - 1]
low = min(prices[i], lowest)
else:
dp[i] = max(dp[i - 1], prices[i] - low)
return dp[-1]

greedy:

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class Solution:
def maxProfit(self, prices: List[int]) -> int:
if not prices:
return 0
low = prices[0]
p = 0
for i in range(1, len(prices)):
p = max(p, prices[i] - low)
low = min(low, prices[i])
return p

高频,三刷

122. Best Time to Buy and Sell Stock II (Easy)

Say you have an array for which the ith element is the price of a given stock on day i.
Design an algorithm to find the maximum profit. You may complete as many transactions as you like (i.e., buy one and sell one share of the stock multiple times).
Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).

Example 1:
Input: [7,1,5,3,6,4]
Output: 7
Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.

Example 2:
Input: [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
engaging multiple transactions at the same time. You must sell before buying again.
Example 3:
Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

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class Solution:
def maxProfit(self, prices: List[int]) -> int:
if not prices:
return 0
n = len(prices)
dp = [0] * n
for i in range(1, n):
if prices[i] > prices[i - 1]:
dp[i] = dp[i - 1] + prices[i] - prices[i - 1]
else:
dp[i] = dp[i - 1]
return dp[-1]

高频,二刷:可以无限买卖=可以抓住所有价格上升的机会,可优化为O(1)空间的算法,因为不在乎中间利润,中间利润累加上去返回

91. Decode Ways (Medium)

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A message containing letters from A-Z is being encoded to numbers using the following mapping:

'A' -> 1
'B' -> 2
...
'Z' -> 26
Given a non-empty string containing only digits, determine the total number of ways to decode it.

Example 1:

Input: "12"
Output: 2
Explanation: It could be decoded as "AB" (1 2) or "L" (12).
Example 2:

Input: "226"
Output: 3
Explanation: It could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).

高频

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class Solution:
def numDecodings(self, s: str) -> int:
if s[0] == "0":
return 0
n = len(s)
f = [0] * (n + 1)
f[0] = f[1] = 1
for i in range(2, n + 1):
one = int(s[i - 1: i])
two = int(s[i - 2: i])
if one > 0:
f[i] += f[i - 1]
if 10 <= two <= 26:
f[i] += f[i - 2]
return f[n]

总结:f[n]为第n个数的方案数,0个字1个方案,1个字1个方案,…if 10 <= two <= 26:…防止误判“05”, “06”等情况

62. Unique Paths (Medium)

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A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).

How many possible unique paths are there?

Above is a 7 x 3 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

Example 1:

Input: m = 3, n = 2
Output: 3
Explanation:
From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
1. Right -> Right -> Down
2. Right -> Down -> Right
3. Down -> Right -> Right
Example 2:

Input: m = 7, n = 3
Output: 28
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class Solution:
def uniquePaths(self, m: int, n: int) -> int:
if m == 1 or n == 1:
return 1
f = [[1 for _ in range(m + 1)] for _ in range(n + 1)]
for r in range(2, n + 1):
for c in range(2, m + 1):
f[r][c] = f[r - 1][c] + f[r][c - 1]
return f[n][m]

高频

53. Maximum Subarray (Easy)

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Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.

Example:

Input: [-2,1,-3,4,-1,2,1,-5,4],
Output: 6
Explanation: [4,-1,2,1] has the largest sum = 6.
Follow up:

If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.
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class Solution:
def maxSubArray(self, nums: List[int]) -> int:
dp = [0] * len(nums)
dp[0] = nums[0]
for i in range(1, len(nums)):
if dp[i - 1] < 0:
dp[i] = nums[i]
else:
dp[i] = dp[i - 1] + nums[i]
return max(dp)

高频,面经:维萨…if dp[i - 1] < 0:…

63. Unique Paths II (Medium)

A robot is located at the top-left corner of a m x n grid (marked ‘Start’ in the diagram below).
The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked ‘Finish’ in the diagram below).
Now consider if some obstacles are added to the grids. How many unique paths would there be?
An obstacle and empty space is marked as 1 and 0 respectively in the grid.
Note: m and n will be at most 100.

Example 1:
Input:
[
[0,0,0],
[0,1,0],
[0,0,0]
]
Output: 2
Explanation:
There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:

  1. Right -> Right -> Down -> Down
  2. Down -> Down -> Right -> Right

space O(mxn):

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class Solution:
def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
m, n = len(obstacleGrid), len(obstacleGrid[0])
dp = [[0] * n for _ in range(m)]
obs = False
for i, v in enumerate(obstacleGrid[0]):
if v == 1:
obs = True
if obs:
dp[0][i] = 0
else:
dp[0][i] = 1
obs = False
for i in range(m):
if obstacleGrid[i][0] == 1:
obs = True
if obs:
dp[i][0] = 0
else:
dp[i][0] = 1
for r in range(1, m):
for c in range(1, n):
if obstacleGrid[r][c] == 1:
dp[r][c] = 0
else:
dp[r][c] = dp[r - 1][c] + dp[r][c - 1]
return dp[m - 1][n - 1]

space: O(n):

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class Solution:
def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
m, n = len(obstacleGrid), len(obstacleGrid[0])
dp = [0] * n
dp[0] = 1 if obstacleGrid[0][0] == 0 else 0
obs = False
for i in range(1, n):
if obstacleGrid[0][i] == 1:
obs = True
if obs:
dp[i] = 0
else:
dp[i] = dp[i - 1]
for r in range(1, m):
dp[0] = dp[0] if obstacleGrid[r][0] == 0 else 0
for c in range(1, n):
if obstacleGrid[r][c] == 1:
dp[c] = 0
else:
dp[c] = dp[c - 1] + dp[c]
return dp[n - 1]

高频
面经:Celo, todo: space: O(1)

64. Minimum Path Sum (Medium)

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.

Note: You can only move either down or right at any point in time.

Example:

Input:
[
[1,3,1],
[1,5,1],
[4,2,1]
]
Output: 7
Explanation: Because the path 1→3→1→1→1 minimizes the sum.
空间O(m*n):

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class Solution:
def minPathSum(self, grid: List[List[int]]) -> int:
rl = len(grid)
cl = len(grid[0])
f = [[0 for _ in range(cl)] for _ in range(rl)]
f[0][0] = grid[0][0]
for r in range(1, rl):
f[r][0] = f[r - 1][0] + grid[r][0]
for c in range(1, cl):
f[0][c] = f[0][c - 1] + grid[0][c]
for r in range(1, rl):
for c in range(1, cl):
f[r][c] = min(f[r - 1][c], f[r][c - 1]) + grid[r][c]
return f[rl - 1][cl - 1]

空间O(n):

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class Solution:
def minPathSum(self, grid: List[List[int]]) -> int:
m, n = len(grid), len(grid[0])
dp = [0] * n
dp[0] = grid[0][0]
for c in range(1, n):
dp[c] = dp[c - 1] + grid[0][c]
for r in range(1, m):
dp[0] += grid[r][0]
for c in range(1, n):
dp[c] = min(dp[c], dp[c - 1]) + grid[r][c]
return dp[-1]

高频:优化空间可以将二维数组f压缩为一维,因为答案不关心中间值
面经:Amazon。优化空间

72. Edit Distance (Hard)

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Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.

You have the following 3 operations permitted on a word:

Insert a character
Delete a character
Replace a character
Example 1:

Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation:
horse -> rorse (replace 'h' with 'r')
rorse -> rose (remove 'r')
rose -> ros (remove 'e')
Example 2:

Input: word1 = "intention", word2 = "execution"
Output: 5
Explanation:
intention -> inention (remove 't')
inention -> enention (replace 'i' with 'e')
enention -> exention (replace 'n' with 'x')
exention -> exection (replace 'n' with 'c')
exection -> execution (insert 'u')
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高频:f 为从 m 的前 i 个变到 n 的前 j 个字符串所需的最少步骤, 第前 i 个字符串 == word1[i - 1]…if word1[i - 1] == word2[j - 1]: f[i][j] = f[i - 1][j - 1]…else: f[i][j] = min(f[i - 1][j - 1], f[i - 1][j], f[i][j - 1]) + 1

132. Palindrome Partitioning II (Hard)

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Given a string s, partition s such that every substring of the partition is a palindrome.

Return the minimum cuts needed for a palindrome partitioning of s.

Example:

Input: "aab"
Output: 1
Explanation: The palindrome partitioning ["aa","b"] could be produced using 1 cut.
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class Solution: