【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 = ; i < n; ++i)
             A[] ^= A[i];
         return A[];
     }
 };

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。