Subsets II

//90. Subsets II

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],
[]
]

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

class Solution {
public:
    vector<vector<int> > subsetsWithDup(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        res.clear();
        if(nums.size()==0){
            return res;
        }
        vector<int> s;
        for(int i=0;i<nums.size();i++){
            findNext(nums,0,i,s);
        }
        res.push_back(nums);
        return res;
    }
private:
    vector<vector<int> > res;
    //s saved 0---start-1,nums 
    void findNext(vector<int>& nums,int start,int k,vector<int>& s){
        if(s.size()==k){
            res.push_back(s);
            return;
        }

        for(int i=start;i<nums.size();i++){
            //deal with the i element
            s.push_back(nums[i]);
            //dela with the i+1 element

            findNext(nums,i+1,k,s);
            s.pop_back();
            //recursuion end,when to push next,skip same value first
            while(i<nums.size()-1 && nums[i]==nums[i+1] ){
                i++;
            }
        }
        return;
    }
};

int main(){
    int arr[]={4,4,4,1,4};
    vector<int> nums(arr,arr+sizeof(arr)/sizeof(int));
    vector<vector<int> > res=Solution().subsetsWithDup(nums);
    for(int i=0;i<res.size();i++){
        for(int j=0;j<res[i].size();j++){
            cout<<res[i][j]<<" ";
        }
        cout<<endl;
    }
    return 0;
}