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 = 0;
        for(std::size_t i=0;i<nums.size();i++)
        {
            std::size_t size = res.size();
            for(std::size_t j=0;j<size;j++)
            {
                if(i== 0 || nums[i] != nums[i-1] || 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, 0, 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() == 0 || nums[start] != v[v.size()-1])
        {
            subsetsWithDupCore(res, start+1, nums, v);    
        }
        
        v.push_back(nums[start]);
        subsetsWithDupCore(res, start+1, nums, v);
        v.pop_back();
    }
};