LeetCode_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],

  []

]

DFS ：

class Solution {

public:

    void  DFS(vector<int> &S, vector<int> &temp,int n, int size,int start)

    {

    

        if(n == size)

        {

            result.push_back(temp);

            return ;

        }

        if(n > size)

            return ;

        

        for(int i = start; i< len ;i++)

        {

            if(i != start && S[i] == S[i-1]) 

                  continue ;

            if(flag[i] == false)

            {

                flag[i] = true;

                temp.push_back(S[i]);

                DFS(S, temp, n+1, size,i+1);

                temp.pop_back();

                flag[i] = false;

            }

        }

    

    }

    vector<vector<int> > subsetsWithDup(vector<int> &S) {

        // Start typing your C/C++ solution below

        // DO NOT write int main() function

        result.clear();

        len = S.size();

        flag.resize(len,false);

        vector<int> temp;

        result.push_back(temp) ;

        sort(S.begin(), S.end());

        

        for(int i = 1; i <= len ; i++)

             DFS(S, temp,0, i,0);

             

        return result;

    }

private:

    vector<vector<int> > result ;

    vector<bool> flag;

    int len;

};

 解释下这句：”if(i != start && S[i] == S[i-1]) continue ; 这句话保证每个循环进入的元素不会有重复。start ： 保证所有的元素进入的顺序为从左往右~

 

重写后的代码：

class Solution {

public:

void DFS(vector<int> &S, vector<int> &ans, int size, int currentPos)

    {

    

        if(ans.size() == size ){

            res.push_back(ans);

            return;

        }

        

        for(int i = currentPos; i< S.size(); ++i)

        {

            if(i != currentPos && S[i] == S[i-1]) continue;    

            ans.push_back(S[i]);

            DFS(S, ans, size, i+1);

            ans.pop_back();

        }

    }

    vector<vector<int> > subsetsWithDup(vector<int> &S) {

        // Start typing your C/C++ solution below

        // DO NOT write int main() function

        res.clear();

        sort(S.begin(), S.end());

        vector<int> emp;

        res.push_back(emp);

        

        for(int i = 1; i <= S.size(); ++i)

        {

            vector<int> ans;

            DFS(S, ans, i, 0);

        }

        

        return res;

    }

private:

    vector<vector<int>> res;

};