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?
public class Solution { public int singleNumber(int[] nums) { //一个数和他本身异或，为0 int res=0; for(int i=0;i<nums.length;i++){ res^=nums[i]; } return res; } }
原文：http://www.cnblogs.com/qiaomu/p/4678131.html