LeetCode_Permutations II

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

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;

};