leetcode:172. factorial trailing zeroes

题目概述

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

Example 1:

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Input: 3
Output: 0
Explanation: 3! = 6, no trailing zero.

Example 2:

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Input: 5
Output: 1
Explanation: 5! = 120, one trailing zero.

Note: Your solution should be in logarithmic time complexity.

在阶乘中，只有5的倍数能产生0，因此这道题就转化成了n!的因子中有多少个5.

这里需要注意的是，25，125这种数字包含了多个5，需要进行重复计算。

代码实现

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class :
    def trailingZeroes(self, n: int) -> int:
        divisor = 5
        result = 0
        while divisor <= n:
            result += n // divisor
            divisor *= 5
        return result