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? 

o(n)的算法只能是线性扫描一遍，可能的相法是位运算。对于异或来说：

1. 异或运算是可交换，即 a ^ b = b ^ a

2. 0 ^ a = a

那么如果对所有元素做异或运算，其结果为那个出现一次的元素，理解是a1 ^ a2 ^ ....，可以将所有相同元素交换至相邻位置，首先运算相同元素，则会产生(n - 1)/2个0异或积，剩余一个单一元素，他们的异或积为这个单一元素自己，得解。

public class Solution {

	public int singleNumber(int[] A) {
		if (A == null || A.length == 0) {
			return 0;
		}
		int result = A[0];
		for (int i = 1; i < A.length; i++) {
			result = result ^ A[i];
		}
		return result;
	}
}