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Given a set of distinct integers, nums, return all possible subsets.
Note:
For example,
If nums = [1,2,3], a solution is:
[ [3], [1], [2], [1,2,3], [1,3], [2,3], [1,2], [] ]
class Solution { public: void dfs(vector<vector<int>>&u,int x,vector<int>& nums,vector<int>&v,int cnt,int n){ if(cnt==n){ //n++; u.push_back(v); return ; } for(int i=x;i<nums.size();i++){ v.push_back(nums[i]); dfs(u,i+1,nums,v,cnt+1,n); v.pop_back(); } } vector<vector<int>> subsets(vector<int>& nums) { vector<vector<int>>u; vector<int>v; sort(nums.begin(),nums.end()); for(int i=0;i<=nums.size();i++){ dfs(u,0,nums,v,0,i); } return u; } };
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原文地址：http://www.cnblogs.com/pk28/p/5356236.html
