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?

 

在数组中，数字皆成对出现，找出唯一一个单独的数字。

思路：a^a=0,相同数字异或为0.

 

1 public class Solution { 2     public int singleNumber(int[] A) { 3         if(A==null||A.length==0) 4             return 0; 5         int res=0; 6         for(int i=0;i