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;
};