<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>

const static auto io_speed_up = []() {
    std::ios::sync_with_stdio(0);
    cin.tie(0);
    FILE *fptr = fopen("user.out", "w");
    int x, out = 0;
    while (!cin.eof()) {
        if (cin.peek() == '[') cin.ignore();
        else break;
        out = 0;
        while (cin.peek() != ']') {
            cin >> x;
            if (cin.peek() == ',') cin.ignore();
            out ^= x;
        }
        fprintf(fptr, "%d\n", out);
        cin.ignore(1024,'\n');
    }
    fclose(fptr);
    exit(0);
    return 0;
}();

class Solution {
public:
    int singleNumber(vector<int>& nums) { return 0; }
};
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