LeetCode_Subsets II

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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;
};