leetcode - 39. Combination Sum - Medium descrition Given a set of candidate numbers (C) (without duplicates) and a target number (T), find all unique combinations in C where the candidate numbers sums to T. The same repeated number may be chosen from…

Given two sentences words1, words2 (each represented as an array of strings), and a list of similar word pairs pairs, determine if two sentences are similar. For example, words1 = ["great", "acting", "skills"] and words2 = [&…

leetcode - 48. Rotate Image - Medium descrition 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 direct…

leetcode - 31. Next Permutation - Medium descrition Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers. If such arrangement is not possible, it must rearrange it as the lowest possible…

[题目] 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 unlimi…

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…

Description: Given a string of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. The valid operators are+, - and *. Example 1 Input: "2-1-1". ((2-1)-1) = 0 (2-(1-1))…

最先看到这一题,直觉的解法就是len从1到s1.size()-1,递归调用比較s1和s2长度为len的子串是否相等.以及剩余部分是否相等. 将s1分成s1[len + rest],分别比較s2[len + rest]和s2[rest + len] 代码例如以下: bool isScramble(string s1, string s2) { return find(s1, s2); } bool find(string s1, string s2) { if (s1.compare(s2) ==…

代码: #include<iostream> #include<vector> using namespace std; struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {} }; ListNode* rotateRight(ListNode* head, int k) { if (head == NULL) return head; ListNode * p = hea…

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 n…