最初想法

def findErrorNums(nums):n = len(nums)duplicate = -1missing = -1for num in nums:if nums[abs(num) - 1] < 0:duplicate = abs(num)else:nums[abs(num) - 1] *= -1for i in range(n):if nums[i] > 0:missing = i + 1breakreturn [duplicate, missing]

遇到力扣大佬

class Solution:def findErrorNums(self, nums: List[int]) -> List[int]:n=len(nums)S=sum(set(nums))#set()返回列表中独特的数，比如输入[1,2,2,4]会返回[1,2,4]return [sum(nums)-S,n*(n+1)//2-S]#后者代表1-N求和

我擦，被秒了