leetcode 136. Single Number && 137. Single Number II

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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?

异或运算应用：

　　异或运算满足交换律和结合律, 通过交换元素(如插入排序等)可以使得相同元素相邻，

(x, x), (y, y), ..., sigle, ... , (z, z)

所有元素异或后：(x ^ x) ^ ... ^ single ^ (z ^ z) = 0 ^ ... ^ single ^ ... ^ 0 = single 

1 int singleNumber(int A[], int n) 
2     {
3         int single = 0;
4         
5         for (int i = 0; i < n; i++)
6             single ^= A[i];
7         
8         return single; 
9     }

leetcode 136. Single Number && 137. Single Number II

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原文地址：http://www.cnblogs.com/ym65536/p/4230658.html