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+<h2><a href="https://leetcode.com/problems/power-of-two/">231. Power of Two</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 two. Otherwise, return <code>false</code></em>.</p>
+
+<p>An integer <code>n</code> is a power of two, if there exists an integer <code>x</code> such that <code>n == 2<sup>x</sup></code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre><strong>Input:</strong> n = 1
+<strong>Output:</strong> true
+<strong>Explanation: </strong>2<sup>0</sup> = 1
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre><strong>Input:</strong> n = 16
+<strong>Output:</strong> true
+<strong>Explanation: </strong>2<sup>4</sup> = 16
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre><strong>Input:</strong> n = 3
+<strong>Output:</strong> false
+</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>
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