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<h2><a href="https://leetcode.com/problems/power-of-three/">326. Power of Three</a></h2><h3>Easy</h3><hr><div><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>
<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>
<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> n = 27
<strong>Output:</strong> true
<strong>Explanation:</strong> 27 = 3<sup>3</sup>
</pre>
<p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> n = 0
<strong>Output:</strong> false
<strong>Explanation:</strong> There is no x where 3<sup>x</sup> = 0.
</pre>
<p><strong class="example">Example 3:</strong></p>
<pre><strong>Input:</strong> n = -1
<strong>Output:</strong> false
<strong>Explanation:</strong> There is no x where 3<sup>x</sup> = (-1).
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>-2<sup>31</sup> &lt;= n &lt;= 2<sup>31</sup> - 1</code></li>
</ul>
<p>&nbsp;</p>
<strong>Follow up:</strong> Could you solve it without loops/recursion?</div>