# 阶乘后的零

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

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

<p>&nbsp;</p>

<p><strong>示例 1：</strong></p>

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

<p><strong>示例 2：</strong></p>

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

<p><strong>示例 3：</strong></p>

<pre>
<strong>输入：</strong>n = 0
<strong>输出：</strong>0
</pre>

<p>&nbsp;</p>

<p><strong>提示：</strong></p>

<ul>
	<li><code>0 &lt;= n &lt;= 10<sup>4</sup></code></li>
</ul>

<p>&nbsp;</p>

<p><b>进阶：</b>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？</p>


## template

```python
class Solution:
    def trailingZeroes(self, n: int) -> int:
        zero_count = 0
        current_multiple = 5
        while n >= current_multiple:
            zero_count += n // current_multiple
            current_multiple *= 5
        return zero_count

```

## 答案

```python

```

## 选项

### A

```python

```

### B

```python

```

### C

```python

```