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?
给定一个数组,其中只有一个数字出现了1次,其他的数字都出现了2次。求出这个出现一次的数字。这道题利用XOR进行求解,如果两个数字相同,那么它们的XOR为0,然而0与任何数XOR都是任何数。所以用0与数组中各个数XOR,最后的结果就是那个只出现一次的数字。
class Solution { public: int singleNumber(vector<int>& nums) { int res = 0; for (int num : nums) res ^= num; return res; } }; // 13 ms