leetcode-problemset/leetcode-cn/originData/factorial-trailing-zeroes.json

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            "title": "Factorial Trailing Zeroes",
            "titleSlug": "factorial-trailing-zeroes",
            "content": "<p>Given an integer <code>n</code>, return <em>the number of trailing zeroes in </em><code>n!</code>.</p>\n\n<p>Note that <code>n! = n * (n - 1) * (n - 2) * ... * 3 * 2 * 1</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 3\n<strong>Output:</strong> 0\n<strong>Explanation:</strong> 3! = 6, no trailing zero.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 5\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> 5! = 120, one trailing zero.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 0\n<strong>Output:</strong> 0\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>0 &lt;= n &lt;= 10<sup>4</sup></code></li>\n</ul>\n\n<p>&nbsp;</p>\n<p><strong>Follow up:</strong> Could you write a solution that works in logarithmic time complexity?</p>\n",
            "translatedTitle": "阶乘后的零",
            "translatedContent": "<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>\n\n<p>提示&nbsp;<code>n! = n * (n - 1) * (n - 2) * ... * 3 * 2 * 1</code></p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 3\n<strong>输出：</strong>0\n<strong>解释：</strong>3! = 6 ，不含尾随 0\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 5\n<strong>输出：</strong>1\n<strong>解释：</strong>5! = 120 ，有一个尾随 0\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 0\n<strong>输出：</strong>0\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>0 &lt;= n &lt;= 10<sup>4</sup></code></li>\n</ul>\n\n<p>&nbsp;</p>\n\n<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>\n",
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                    "code": "class Solution {\npublic:\n    int trailingZeroes(int n) {\n\n    }\n};",
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