leetcode 第136题 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?

思路：题目要求时间复杂度为O（n）,并且空间复杂度为O（1）.考虑用位操作。这里我们用异或^。
异或规则：如果两个比特位相同，异或之后变为0，两个比特位不同，则异或之后为1.且a ^ b = b ^ a, a ^ 0 = a.
我们将数组中的数从头到尾异或一遍，得到的值即为所求。

C++代码实现：

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