该文章创建(更新)于03/24/2022，请注意文章的时效性！

Given an integer n, return the number of trailing zeroes in n!.

Note that n! = n * (n - 1) * (n - 2) * ... * 3 * 2 * 1.

• https://leetcode.com/problems/factorial-trailing-zeroes/

Solution

Count How many 5 in the n

Code

• python

class Solution:
    def trailingZeroes(self, n: int) -> int:
        ret = 0
        while n >= 5:
            n //= 5
            ret += n
        return ret

要不赞赏一下?

微信

支付宝

PayPal

Bitcoin