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

 class Solution {
 public:
     vector<vector<int>> subsetsWithDup(vector<int>& nums) {
         sort(nums.begin(), nums.end());
         vector<int> v;
         vector<vector<int>> res;

         res.push_back(v); //empty set

         std::size_t prev_size = ;
         for(std::size_t i=;i<nums.size();i++)
         {
             std::size_t size = res.size();
             for(std::size_t j=;j<size;j++)
             {
                 if(i==  || nums[i] != nums[i-] || j >= prev_size)
                 {
                     v.assign(res[j].begin(),res[j].end());
                     v.push_back(nums[i]);
                     res.push_back(v);
                 }
             }
             prev_size = size;
         }
         return res;
     }
 };

方法二：递归

 #include <vector>
 #include <algorithm>
 using std::vector;

 class Solution {
 public:
     vector<vector<int>> subsetsWithDup(vector<int>& nums) {
         sort(nums.begin(), nums.end());
         vector<int> v;
         vector<vector<int>> res;
         subsetsWithDupCore(res, , nums, v);
         return res;
     }
 private:
     void subsetsWithDupCore(vector<vector<int>>& res,int start,vector<int>& nums,vector<int>& v)
     {
         if(start == nums.size())
         {
             res.push_back(v);
             return;
         }
         if(v.size() ==  || nums[start] != v[v.size()-])
         {
             subsetsWithDupCore(res, start+, nums, v);
         }

         v.push_back(nums[start]);
         subsetsWithDupCore(res, start+, nums, v);
         v.pop_back();
     }
 };