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?

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

异或在这里就完美演绎了，所有数字在计算机都是0和1，相同的都将消失，留下的就是那个Single Number，我很喜欢这道题。