factorial-trailing-zeroes

java

python

python3

mysql

mssql

oraclesql

csharp

javascript

typescript

bash

swift

kotlin

dart

golang

ruby

scala

rust

racket

erlang

elixir

pythondata

postgresql

cangjie

Algorithms

Given an integer <code>n</code>, return the number of trailing zeroes in <code>n!</code>.

Note that <code>n! = n * (n - 1) * (n - 2) * ... * 3 * 2 * 1</code>.

&nbsp;
Example 1:

<pre>
Input: n = 3
Output: 0
Explanation: 3! = 6, no trailing zero.
</pre>

Example 2:

<pre>
Input: n = 5
Output: 1
Explanation: 5! = 120, one trailing zero.
</pre>

Example 3:

<pre>
Input: n = 0
Output: 0
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>0 &lt;= n &lt;= 104</code></li>
</ul>

&nbsp;
Follow up: Could you write a solution that works in logarithmic time complexity?

trailingZeroes

Factorial Trailing Zeroes

Number of Digit One

number-of-digit-one

数字 1 的个数

Preimage Size of Factorial Zeroes Function

preimage-size-of-factorial-zeroes-function

阶乘函数后 K 个零

Abbreviating the Product of a Range

abbreviating-the-product-of-a-range

一个区间内所有数乘积的缩写

Maximum Trailing Zeros in a Cornered Path

maximum-trailing-zeros-in-a-cornered-path

转角路径的乘积中最多能有几个尾随零

Math

给定一个整数 <code>n</code> ，返回 <code>n!</code> 结果中尾随零的数量。

提示&nbsp;<code>n! = n * (n - 1) * (n - 2) * ... * 3 * 2 * 1</code>

&nbsp;

示例 1：

<pre>
输入：n = 3
输出：0
解释：3! = 6 ，不含尾随 0
</pre>

示例 2：

<pre>
输入：n = 5
输出：1
解释：5! = 120 ，有一个尾随 0
</pre>

示例 3：

<pre>
输入：n = 0
输出：0
</pre>

&nbsp;

提示：

<ul>
	<li><code>0 &lt;= n &lt;= 104</code></li>
</ul>

&nbsp;

进阶：你可以设计并实现对数时间复杂度的算法来解决此问题吗？