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^b^b=a a抑或同一个数两次认为a。

 1 class Solution { 2 public: 3     int singleNumber(int A[], int n) { 4         int res; 5         for(int i=0;i<n;i++) 6         { 7             res=res^A[i]; 8         } 9         return res;10     }11 };