LeetCode——Maximum Depth of Binary Tree Question 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. Answer /*** Definition for a binary tree node

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. For example:Given the below binary tree andsum = 22, 5/ \4 8/ / \11 13 4/ \ / \7 2 5 1 return [[5,4,11,2],[5,8,4,5] ]題意：給定一數，

題目如下： Python代碼： def isPalindrome(self, x):""":type x: int:rtype: bool"""if(x>=0):s = int(str(x)[::-1])if(x==s):return Trueelse:return Falseelse:return False leetcode 53？? 轉載于:https://www.c

Given a singly linked list?L:?L0→L1→…→Ln-1→Ln,reorder it to:?L0→Ln→L1→Ln-1→L2→Ln-2→… You must do this in-place without altering the nodes' values. For example,Given?{1,2,3,4}, reorder it to?{1,4,2,3}. ? LeetCode、OH! MY GOD! I HATE LINKED

給定一個只包含數字的字符串，復原它并返回所有可能的 IP 地址格式。 LeetCode、示例: 輸入: "25525511135" 輸出: ["255.255.11.135", "255.255.111.35"] 注：這里是IPV4,用2進制表示的話要32位，用10進制則是4個整數,每個整數大于

class Solution {int n;vector<string>res;unordered_map<int,set<int>>hash;string s;void dfs(int start, string temp){if(start == n){res.push_back(temp);return ;}for(auto end : hash[start]){if(start == 0){//前面不要有空格dfs

力扣 思路:過程模擬 第一步 矩陣轉置 LeetCode？第二步 矩陣矩陣每行逆置 代碼 ? class Solution { public:void zhuanzhi(vector<vector<int>>&matrix){for(int i=0;i<matrix.size();i++){for(int j=0;j<i;j++){swap(matrix[i

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. 思考：求二叉樹高的變形，加上

Maximum Subarray Find the contiguous subarray within an array (containing at least one number) which has the largest sum. For example, given the array?[?2,1,?3,4,?1,2,1,?5,4],the contiguous subarray?[4,?1,2,1]?has the largest sum =?6. More practice:

Description: Implement pow(x,?n). 分析： 求冪次運算，典型的分治算法來解。 因為pow(x,n/2)*pow(x,n/2) 有著重復運算，分治法就會非常快O(log n) 1 class Solution { 2 public: 3 double findval(double x,int n) 4 { 5 if(n==0) return 1; 6 i