LeetCode_Permutations II

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Given a collection of numbers that might contain duplicates, return all possible unique permutations.

 

For example,

[1,1,2] have the following unique permutations:

[1,1,2], [1,2,1], and [2,1,1].

　　ＤＦＳ　＋　剪枝

class Solution {
public:
   void DFS(vector<int> &num, int size,vector<int> temp)
    {
        if(size == n){
            result.push_back(temp);
            return ;
        }
        for(int i = 0; i< n; i++)
        {
            if(flag[i] || (i!= 0 &&flag[i-1] && num[i] == num[i-1] ) )
                  continue;
            
            temp[size] = num[i];
            flag[i] = true;
            DFS(num, size+1, temp);
            flag[i] = false;
           
        }
    }
    vector<vector<int> > permuteUnique(vector<int> &num) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        n = num.size();
        result.clear();
        
        
        sort(num.begin(), num.end());
        
        if(n == 0) return result;
        flag.resize(n,false);
        vector<int> temp(n,0) ;         
        DFS(num,0,temp);   
        
        return result ;
    }
private :
   int  n;
   vector<bool> flag;
   vector<vector<int>> result;   
};

这里解释下剪枝的原理：　有重复元素的时候，因为重复的元素处理无序所以导致重复，所以只要给重复的元素进入temp定义一个次序就可以去掉重复。这里定义次序的规则是： 先对所有元素排序对于有重复的元素，必须是排在后面的元素比排在前面的元素先进入temp

重写后，貌似比第一个版本要快一点

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class Solution {

public:

    void DFS(vector<int> &num, vector<int> &tp, vector<bool> flag)

    {

        if(num.size() == tp.size()){

            res.push_back(tp);

            return;

        }

        for(int i = 0; i< num.size(); i++)

        {

            if(flag[i] == true) continue;

            if(i != 0 && num[i] == num[i-1] && flag[i-1] == false) continue;

         

            flag[i] = true;

            tp.push_back(num[i]);

            DFS(num, tp, flag);

            tp.pop_back();

            flag[i] = false;

        }

    }

    vector<vector<int> > permuteUnique(vector<int> &num) {

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

        // DO NOT write int main() function

        res.clear();

        int len = num.size();

        if(len < 1) return res;

        sort(num.begin(), num.end());

        vector<bool> flag(len, false);

        vector<int> tp;

        DFS(num, tp, flag);

        return res;

         

    }

private:

    vector<vector<int>> res;

};