算法描述：

Given a set of distinct integers, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

Example:

Input: nums = [1,2,3]Output:[  [3],  [1],  [2],  [1,2,3],  [1,3],  [2,3],  [1,2],  []]

解题思路：题目要求所有可能的子集，因此首先想到回溯法求解。注意边界，以及返回问题。

vector
     
      
       > subsets(vector
       
        & nums) {        vector
        
         
          > results;        vector
          
            temp; backtracking(nums,results, temp, 0); return results; } void backtracking(vector
           
            & nums, vector
            
             
              >& results, vector
              
               & temp, int index){ if(temp.size() <= nums.size()){ results.push_back(temp); } if(temp.size()==nums.size()) return; for(int i=index; i < nums.size();i++){ temp.push_back(nums[i]); backtracking(nums, results, temp, i+1); temp.pop_back(); } }