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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?
解题思路：
按位异或运算，出现两次的都归零了，剩下的就是出现一次的了，JAVA实现如下：
public int singleNumber(int[] nums) {
for(int i=1;i<nums.length;i++)
nums[0]^=nums[i];
return nums[0];
}
Java for LeetCode 136 Single Number
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原文地址：http://www.cnblogs.com/tonyluis/p/4548790.html
