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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], [] ]
1 class Solution { 2 public: 3 vector<vector<int>> subsetsWithDup(vector<int>& nums) { 4 sort(nums.begin(),nums.end()); 5 //无重复 6 int tag = 0; 7 int record = 0; 8 vector<vector<int>> result; 9 vector<int> line; 10 result.push_back(line); 11 line.push_back(nums[0]); 12 result.push_back(line); 13 for(int i = 1;i<nums.size();i++){ 14 if(nums[i]!=nums[i-1]){ 15 int n = result.size(); 16 for(int j = 0; j<n ; j++){ 17 vector<int> temp = result[j]; 18 temp.push_back(nums[i]); 19 result.push_back(temp); 20 tag = 0; 21 } 22 record = n; 23 } 24 if(nums[i]==nums[i-1]){ 25 int n = result.size(); 26 if(tag == 0){//上一个没有重复 27 for(int j = n/2;j<n;j++){ 28 vector<int> temp = result[j]; 29 temp.push_back(nums[i]); 30 result.push_back(temp); 31 tag = 1; 32 } 33 record = n/2; 34 } 35 else if(tag == 1){ 36 for(int j = n-record;j<n;j++){ 37 vector<int> temp = result[j]; 38 temp.push_back(nums[i]); 39 result.push_back(temp); 40 tag = 1; 41 } 42 record = record; 43 } 44 45 } 46 } 47 return result; 48 } 49 };
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原文地址：http://www.cnblogs.com/timesdaughter/p/5564141.html
