[Leetcode 136, Medium] Single Number I

Problem:

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?

Analysis:

算法：逻辑运算的异或（^）是交换且可结合的，而且，是幂等的。所以，如果是有且只有一个单独出现的元素，那么把所有的元素进行异或操作，所得到的值就是所要的结果。

衍生问题：every element appears even times except for one which appears odd times.

Solution: 

C++:

(I)

    int singleNumber(int A[], int n) {
        if(n == 0)
            return 0;

        int rval = A[0];
        for(int i = 1; i < n; ++i)
            rval ^= A[i];
        
        return rval;
    }