C++异或

原题：

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

Note: 
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

代码：

class Solution {
public:
    int singleNumber(int A[], int n) {
        int num = 0;
        for(int  i = 0;i<n; i++){
            num ^= A[i];
        }
        return num;
    }
};

思路：

其实只要记住：

        1.异或满足交换律。

        2.相同两个数异或为0。

        3.0异或一个数为那个数本身。

最后结果即出现1次的那个数。

例子:1^2^3^4^4^3^2 = 2^2^3^3^4^4^1 = 1

拓展：http://blog.youkuaiyun.com/smile_watermelon/article/details/47748227