感谢：http://www.cnblogs.com/changchengxiao/p/3413294.html
Single Number
Total Accepted: 103007 Total Submissions: 217483 Difficulty: Medium
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?
琢磨了很久没找到满足题目的两个要求：O(1)和不开空间。实在没办法才上网查找，原来是汇编语言的知识，而且C++中还可以直接用。
原文：
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： . 异或运算是可交换，即 a ^ b = b ^ a . ^ 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 ^ ....
所以说，这是终极解法。
public class Solution {
public int singleNumber(int[] A) {
// Note: The Solution object is instantiated only once and is reused by each test case.
if(A == null || A.length == ){
return ;
}
int result = A[];
for(int i = ; i < A.length; i++){
result ^= A[i];
}
return result;
}
}
LeetCode很有趣，希望明天可以自己想出来！