# 问题 原题链接：https://leetcode.com/problems/single-number/ Given a non-empty 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? Example 1: ``` Input: [2,2,1] Output: 1 ``` Example 2: ``` Input: [4,1,2,1,2] Output: 4 ``` # 分析过程 - 输入：一个整数数组，所有数字都出现了两次，只有一个数字出现了一次。 - 输出：找到这个只出现了一次的数。 - 思路：任何数，自己按位异或自己都是 0，因此只要循环异或，剩下的数就是要找的数。 # 解决方法 ```cpp class Solution { public: int singleNumber(vector<int>& nums) { int result = 0; for (int num : nums) { result ^= num; } return result; } }; ```