78. Subsets

Given a set of distinct integers, nums, return all possible subsets.

Note: The solution set must not contain duplicate subsets.

For example,
If nums = [1,2,3], a solution is:

[
  [3],
  [1],
  [2],
  [1,2,3],
  [1,3],
  [2,3],
  [1,2],
  []
]
解题思路：递归回溯法
每次在现有的集合下选择加入下一个元素，并把每次的过程都放在最终的result中，然后把下一个元素pop出来，在此情况下选择下下个元素。

class Solution {
public:
    void helper(vector<vector<int>>&res, vector<int>& nums,vector<int> temp, int depth){
        res.push_back(temp);
        for(int i=depth;i<nums.size();i++){
            temp.push_back(nums[i]);
            helper(res,nums,temp,i+1);
            temp.pop_back();
        }
            
    }
    vector<vector<int>> subsets(vector<int>& nums) {
        vector<vector<int>>res;
        vector<int>temp;
        helper(res,nums,temp,0);
        return res;
    }
};