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Description
tab: English
<p>Given an integer <code>n</code>, return <em>the number of trailing zeroes in </em><code>n!</code>.</p>
<p>Note that <code>n! = n * (n - 1) * (n - 2) * ... * 3 * 2 * 1</code>.</p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<pre>
<strong>Input:</strong> n = 3
<strong>Output:</strong> 0
<strong>Explanation:</strong> 3! = 6, no trailing zero.
</pre>
<p><strong class="example">Example 2:</strong></p>
<pre>
<strong>Input:</strong> n = 5
<strong>Output:</strong> 1
<strong>Explanation:</strong> 5! = 120, one trailing zero.
</pre>
<p><strong class="example">Example 3:</strong></p>
<pre>
<strong>Input:</strong> n = 0
<strong>Output:</strong> 0
</pre>
<p> </p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>0 <= n <= 10<sup>4</sup></code></li>
</ul>
<p> </p>
<p><strong>Follow up:</strong> Could you write a solution that works in logarithmic time complexity?</p>
---
[submissions](https://leetcode.com/problems/factorial-trailing-zeroes/submissions/) | [solutions](https://leetcode.com/problems/factorial-trailing-zeroes/solutions/)
tab: 中文
<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 <code>n! = n * (n - 1) * (n - 2) * ... * 3 * 2 * 1</code></p>
<p> </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> </p>
<p><strong>提示:</strong></p>
<ul>
<li><code>0 <= n <= 10<sup>4</sup></code></li>
</ul>
<p> </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>
---
[提交记录](https://leetcode.cn/problems/factorial-trailing-zeroes/submissions/) | [题解](https://leetcode.cn/problems/factorial-trailing-zeroes/solution/)
Solutions & Notes
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