<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>
<pre><strong>Input:</strong> nums = [2,2,1]
<strong>Output:</strong> 1
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> nums = [4,1,2,1,2]
<strong>Output:</strong> 4
</pre><p><strong class="example">Example 3:</strong></p>
<pre><strong>Input:</strong> nums = [1]
<strong>Output:</strong> 1
</pre>
<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>

class Solution:
    def singleNumber(self, nums: List[int]) -> int:
        numTimes = {}
        for i in nums:
            if i not in numTimes:
                numTimes[i] = 1
            else:
                numTimes[i] += 1
        for i in numTimes:
            if numTimes[i] == 1:
                return i
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