题意: 给一个集合,有n个可能相同的元素,求出所有的子集(包括空集,但是不能重复). 思路: 看这个就差不多了.LEETCODE SUBSETS (DFS) class Solution { public: vector<vector<int>> subsets(vector<int>& nums) { sort(nums.begin(),nums.end()); DFS(,nums,tmp); ans.push_back(vector<int>()…

Given a collection of integers that might contain duplicates, S, return all possible subsets. Note: Elements in a subset must be in non-descending order. The solution set must not contain duplicate subsets. For example,If S = [1,2,2], a solution is:…

原题地址:https://oj.leetcode.com/problems/subsets-ii/ 题意: Given a collection of integers that might contain duplicates, S, return all possible subsets. Note: Elements in a subset must be in non-descending order. The solution set must not contain duplicat…

Given a collection of integers that might contain duplicates, S, return all possible subsets. Note: Elements in a subset must be in non-descending order. The solution set must not contain duplicate subsets. For example,If S = [1,2,2], a solution is:…

题目 Given a collection of integers that might contain duplicates, S, return all possible subsets. Note: Elements in a subset must be in non-descending order. The solution set must not contain duplicate subsets. For example, If S = [1,2,2], a solution…

Given a collection of integers that might contain duplicates, S, return all possible subsets. Note: Elements in a subset must be in non-descending order. The solution set must not contain duplicate subsets. For example,If S =[1,2,2], a solution is: […

class Solution {//生成全部[不反复]的组合.生成组合仅仅要採用递归,由序列从前往后遍历就可以. 至于去重,依据分析相应的递归树可知.同一个父节点出来的两个分支不能一样(即不能与前一个元素一样,且前一个元素要与之在同层). public: int *b,n; vector<int>a; vector<vector<int> >ans; void dfs(int id,int len){ if(len>0){ vector<int>v(b…

求集合的所有子集问题 LeetCode:Subsets Given a set of distinct integers, S, return all possible subsets. Note: Elements in a subset must be in non-descending order. The solution set must not contain duplicate subsets. For example,If S = [1,2,3], a solution is:…

目录 题目链接 注意点 解法 小结 题目链接 Subsets II - LeetCode 注意点 有重复的数字 数组可能是无序的,要先排序 解法 解法一:递归,只需要在Subsets中递归写法的基础上多加一句if(find(ret.begin(),ret.end(),tmp) == ret.end()) ret.push_back(tmp);j即可,因为已经排序了,所以加进去的如果已经存在就说明是重复的. class Solution { public: typedef vector<int>…

1. Word Search 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 cel…