subset II

Subsets II

Given a collection of integers that might contain duplicates, nums, return all possible subsets.

Note:

  • Elements in a subset must be in non-descending order.
  • 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],
  []
]
方法一：迭代
 1 class Solution {
 2 public:
 3     vector<vector<int>> subsetsWithDup(vector<int>& nums) {
 4         sort(nums.begin(), nums.end());
 5         vector<int> v;
 6         vector<vector<int>> res;
 7 
 8         res.push_back(v); //empty set
 9 
10         std::size_t prev_size = 0;
11         for(std::size_t i=0;i<nums.size();i++)
12         {
13             std::size_t size = res.size();
14             for(std::size_t j=0;j<size;j++)
15             {
16                 if(i== 0 || nums[i] != nums[i-1] || j >= prev_size)
17                 {
18                     v.assign(res[j].begin(),res[j].end());
19                     v.push_back(nums[i]);
20                     res.push_back(v);
21                 }
22             }
23             prev_size = size;    
24         }
25         return res;
26     }
27 };

 方法二：递归

 1 #include <vector>
 2 #include <algorithm>
 3 using std::vector;
 4 
 5 class Solution {
 6 public:
 7     vector<vector<int>> subsetsWithDup(vector<int>& nums) {
 8         sort(nums.begin(), nums.end());
 9         vector<int> v;
10         vector<vector<int>> res;
11         subsetsWithDupCore(res, 0, nums, v);
12         return res;
13     }
14 private:
15     void subsetsWithDupCore(vector<vector<int>>& res,int start,vector<int>& nums,vector<int>& v)
16     {
17         if(start == nums.size()) 
18         {
19             res.push_back(v);
20             return;
21         }
22         if(v.size() == 0 || nums[start] != v[v.size()-1])
23         {
24             subsetsWithDupCore(res, start+1, nums, v);    
25         }
26         
27         v.push_back(nums[start]);
28         subsetsWithDupCore(res, start+1, nums, v);
29         v.pop_back();
30     }
31 };

 

posted @ 2015-09-28 11:47  wxquare  阅读(230)  评论(0)  编辑  收藏  举报