Power of Two
<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?
Bit Manipulation; Recursion; Math
impl Solution {
    pub fn is_power_of_two(n: i32) -> bool {
        n > 0 && (n & (n - 1)) == 0
    }
}
