[leetcode]Single Number

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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?

算法思路：

抑或^满足交换律和结合律

交换律： a ^ b ^ c ^ d = a ^ d ^ b ^ c;

结合律： a ^ b ^ c ^ d = a ^ （d ^ b） ^ c;

相同数字 a ^ a = 0

代码如下：

1 public class Solution {
2     public int singleNumber(int[] a) {
3         int res = 0;
4         for(int tem : a){
5             res ^= tem;
6         }
7         return res;
8     }
9 }

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[leetcode]Single Number

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原文地址：http://www.cnblogs.com/huntfor/p/3883887.html