【03_136】Single Number

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：

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 ^ ....

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

LeetCode很有趣，希望明天可以自己想出来！