LeetCode 136：Single Number(异或操作)

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?

思路：

因为0^A=A，A^A=0，且异或操作是可交换的，所以该题目使用异或操作就可以找到只出现一次数字。

Code：

class Solution {public:    int singleNumber(vector<int>& nums) {        int result=0;        for(int i=0;i<nums.size();i++) result ^=nums[i];        return result;    }};