【Leetcode】【Medium】Single Number

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?

解题：

此题可直接运用异或运算的性质：N^A^A = N

代码：

class Solution {
public:
    int singleNumber(int A[], int n) {
        for (int i = 1; i < n; ++i) 
            A[0] ^= A[i];
        return A[0];
    }
};

但是在leetcode网站性能分析中，异或运算（19ms）不是最快的，不知道还有没有其他算法可以达到10ms以下。