Problems:
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:这里主要利用的是异或运算的性质，出现偶数次则x清零。基数次保留结果。
class Solution { public: int singleNumber(vector<int>& nums) { //考察的是位运算 int x=0; for(int i=0;i<nums.size();i++) x^=nums[i]; return x; } };
原文：http://www.cnblogs.com/xiaoying1245970347/p/4639254.html