Single Number
<p>Given a <strong>non-empty</strong>&nbsp;array of integers <code>nums</code>, every element appears <em>twice</em> except for one. Find that single one.</p>

<p>You must&nbsp;implement a solution with a linear runtime complexity and use&nbsp;only constant&nbsp;extra space.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [2,2,1]</span></p>

<p><strong>Output:</strong> <span class="example-io">1</span></p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [4,1,2,1,2]</span></p>

<p><strong>Output:</strong> <span class="example-io">4</span></p>
</div>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [1]</span></p>

<p><strong>Output:</strong> <span class="example-io">1</span></p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>-3 * 10<sup>4</sup> &lt;= nums[i] &lt;= 3 * 10<sup>4</sup></code></li>
	<li>Each element in the array appears twice except for one element which appears only once.</li>
</ul>
Bit Manipulation; Array
/**
 * @param {number[]} nums
 * @return {number}
 */
var singleNumber = function (nums) {
    return nums.reduce((a, b) => a ^ b);
};
