把字符串转换成整数 class Solution { public: int StrToInt(string str) { int n = str.size(), s = 1; long long res = 0; if(!n) return 0; if(str[0] == '-') s = -1; for(int i = (str[0] == '-' || str[0] == '+') ? 1 : 0; i < n; ++i){ if(!('0' <= str[i] && s…

leetcode 39 先排序,然后dfs 注意先整全局变量可以减少空间利用 class Solution { vector<vector<int>>ret; vector<int>temp; vector<int> srt; public: vector<vector<int>> combinationSum(vector<int>& candidates, int target) { srt=candidate…

最短路径=>BFS    所有路径=>DFS 126. Word Ladder II BFS+DFS: BFS找出下一个有效的word进队 并记录step 更新两个变量:unordered_map<string, vector<string>> next, unordered_map<string,int> visit DFS找出所有的解法 更新两个变量:vector<vector<string>> result, vector<…

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. Example 1: Input: 1 1 / \ / \ 2 3 2 3 [1,2,3], [1,2,3] Outpu…

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. Example: Given binary tree [3,9,20,null,null,15,7…

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: Given binary tree [3,9,20,null,null,15,7…

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…

代码: 个人浅薄的认为DFS就是回溯法中的一种,一般想到用DFS我们脑中一般都有一颗解法树,然后去按照深度优先搜索去寻找解.而分支界限法则不算是回溯,无论其是采用队列形式的还是优先队列形式的分支界限法. 下面这个函数就是我的DFS的函数,先介绍一下参数的含义,index表示当前要判断是否合适的candidates中的元素的下标,t表示即将要把新数据加入的位置. void backTrace(int sum, int target, int a[], int index, int t, vecto…

BFS, DFS 的题目总结. Directed graph: Directed Graph Loop detection and if not have, path to print all path. BFS/DFS: (可以用BFS或者DFS的,主要还是遍历) [LeetCode] 733. Flood Fill_Easy tag: BFS     1 [LeetCode] 690. Employee Importance_Easy tag: BFS    1 [LeetCode] 529…

参考[LeetCode] questions conlusion_InOrder, PreOrder, PostOrder traversal 可以对binary tree进行遍历. 此处说明Divide and Conquer 的做法,其实跟recursive的做法很像,但是将结果存进array并且输出,最后conquer (这一步worst T:O(n)) 起来,所以时间复杂度可以从遍历O(n) -> O(n^2). 实际上代码是一样, 就是把[root.val] 放在先, 中, 后就是pr…