Subsets II

Given a collection of integers that might contain duplicates, S, 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 S = [1,2,2], a solution is:

[
  [2],
  [1],
  [1,2,2],
  [2,2],
  [1,2],
  []
]

C++代码实现：

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;

class Solution
{
public:
    vector<vector<int> > subsetsWithDup(vector<int> &S)
    {
        if(S.empty())
            return vector<vector<int>>();
        sort(S.begin(),S.end());
        int n=1<<S.size();
        int i=0;
        vector<vector<int> > ret;
        while(i<n)
        {
            vector<int> temp;
            int index=0;
            int j=i;
            while(j>0)
            {
                if(j&1)
                    temp.push_back(S[index]);
                j=j>>1;
                index++;
            }
            int k;
            for(k=0;k<(int)ret.size();k++)
            {
                if(ret[k]==temp)
                    break;
            }
            if(k==(int)ret.size())
                ret.push_back(temp);
            i++;
        }
        return ret;
    }
};

int main()
{
    Solution s;
    vector<int> vec= {2,1,2};
    vector<vector<int> > result=s.subsetsWithDup(vec);
    for(auto a:result)
    {
        for(auto v:a)
            cout<<v<<" ";
        cout<<endl;
    }
}

方法二：

class Solution {
public:
    vector<vector<int> > subsetsWithDup(vector<int> &S) {
        if(S.empty())
            return vector<vector<int> > ();
        sort(S.begin(),S.end());
        vector<vector<int> > res={{}};
        vector<vector<int> > last;
        vector<vector<int> > tmp;
        int i,j;
        int n=S.size();
        int m=res.size();
        for(i=0;i<n;i++)
        {
            if(i>0&&S[i]==S[i-1])
            {
                m=last.size();
                tmp=last;
            }
            else
            {
                m=res.size();
                tmp=res;
            }
            last.clear();
            for(j=0;j<m;j++)
            {
                tmp[j].push_back(S[i]);
                last.push_back(tmp[j]);
                res.push_back(tmp[j]);
            }
        }
        return res;
    }
};

https://leetcode.com/discuss/25834/share-a-clear-solution