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Given a collection of integers that might contain duplicates, nums, return all possible subsets.
Note: The solution set must not contain duplicate subsets.
For example,
If nums = [1,2,2], a solution is:
[ [2], [1], [1,2,2], [2,2], [1,2], [] ]
思路：做法和Leetcode 78. Subsets类似，只是多了重复元素，所以只要去除重复元素的情况就可以。
代码：
1 class Solution { 2 public: 3 void subsetsWithDup(vector<vector<int> >& res,vector<int> v,vector<int> nums,int cur){ 4 res.push_back(v); 5 for(int i=cur;i<nums.size();i++){ 6 v.push_back(nums[i]); 7 subsetsWithDup(res,v,nums,i+1); 8 v.pop_back(); 9 while(i+1<nums.size()&&nums[i]==nums[i+1]) i++; 10 } 11 } 12 vector<vector<int> > subsetsWithDup(vector<int>& nums) { 13 vector<vector<int> > res; 14 vector<int> v; 15 sort(nums.begin(),nums.end()); 16 subsetsWithDup(res,v,nums,0); 17 return res; 18 } 19 };
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原文地址：http://www.cnblogs.com/Deribs4/p/5726625.html
