算法描述：

Given a collection of integers that might contain duplicates, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

Example:

Input: [1,2,2]Output:[  [2],  [1],  [1,2,2],  [2,2],  [1,2],  []]

解题思路：题目要求列出所有可能性，因此用回溯法求解。与前面题目不同，这个题目需要考虑重复元素问题。注意去重是与index开始比较而不是从零开始比较。

vector
     
      
       > subsetsWithDup(vector
       
        & nums) {        sort(nums.begin(),nums.end());        vector
        
         
          > results;        vector
          
            temp; backtracking(results, nums, temp, 0); return results; } void backtracking(vector
           
            
             >& results, vector
             
               nums, vector
              
               & temp, int index){ if(temp.size() <= nums.size()){ results.push_back(temp); }else return; for(int i=index; i 
               
                index && nums[i] == nums[i-1]) continue; temp.push_back(nums[i]); backtracking(results, nums, temp, i+1); temp.pop_back(); } }