最初想法

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