136. 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?

题意：一组数，每个数都会出现两次，只有一个数会出现一次，找到那个唯一的。

思路：异或运算。利用异或的性质，统一都异或一遍，则结果就是唯一的那一个。补充：0^num=num;num^num=0。

class Solution {public:int singleNumber(vector<int>& nums) {if (nums.empty())return 0;int sum = nums[0];for (int i = 1; i < nums.size(); i++){sum ^= nums[i];}return sum;}};