leetcode-problemset/leetcode-cn/originData/power-of-three.json

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            "title": "Power of Three",
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            "content": "<p>Given an integer <code>n</code>, return <em><code>true</code> if it is a power of three. Otherwise, return <code>false</code></em>.</p>\n\n<p>An integer <code>n</code> is a power of three, if there exists an integer <code>x</code> such that <code>n == 3<sup>x</sup></code>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 27\n<strong>Output:</strong> true\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 0\n<strong>Output:</strong> false\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 9\n<strong>Output:</strong> true\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>-2<sup>31</sup> &lt;= n &lt;= 2<sup>31</sup> - 1</code></li>\n</ul>\n\n<p>&nbsp;</p>\n<strong>Follow up:</strong> Could you solve it without loops/recursion?",
            "translatedTitle": "3 的幂",
            "translatedContent": "<p>给定一个整数，写一个函数来判断它是否是 3&nbsp;的幂次方。如果是，返回 <code>true</code> ；否则，返回 <code>false</code> 。</p>\n\n<p>整数 <code>n</code> 是 3 的幂次方需满足：存在整数 <code>x</code> 使得 <code>n == 3<sup>x</sup></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 = 27\n<strong>输出：</strong>true\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 0\n<strong>输出：</strong>false\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 9\n<strong>输出：</strong>true\n</pre>\n\n<p><strong>示例 4：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 45\n<strong>输出：</strong>false\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>-2<sup>31</sup> &lt;= n &lt;= 2<sup>31</sup> - 1</code></li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>进阶：</strong>你能不使用循环或者递归来完成本题吗？</p>\n",
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