leetcode-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：

     a^a=0;  a^0=a;将所有元素进行异或

代码：

int singleNumber(int A[], int n) {
     int res=0;
for(int i=0; i<n; ++i)
{
res=res^A[i];
}
return res;   
    }