Given an array of integers, every element appears twice except for one. Find that single one.

Note:

起初看到题目，不知道怎么做，看了提示说用位运算，马上就有思路了

class Solution {public:	int singleNumber(vector
    
     & nums) {		int ret = 0;		for (int i = 0; i < nums.size(); i++)		{			ret = ret ^ nums[i];		}		return ret;	}};int main(int argc, char *argv[]){	Solution s;	vector
     
       v = { 1, 2, 2, 3,1 };		cout << s.singleNumber(v);	system("pause");	return 0;}
     
    

同样的原理（^异或运算的性质(满足交换律 结合律)）可完成ab的交换而不利用临时变量,许多hash 函数也利用了该性质 

a=a^b;b=a^b;a=a^b;