题目
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?
解题思路
对于异或运算，我们知道 a^a=0; a^0=a，所以我们就把所有的数都异或到一起，那么所有出现两次的数就会是0，只留下最后一个只出现过一次的数字了。
代码
import functools
class Solution:
def singleNumber(self, nums: List[int]) -> int:
return functools.reduce(lambda x, y: x ^ y, nums)
