Merge Intervals
<p>Given an array&nbsp;of <code>intervals</code>&nbsp;where <code>intervals[i] = [start<sub>i</sub>, end<sub>i</sub>]</code>, merge all overlapping intervals, and return <em>an array of the non-overlapping intervals that cover all the intervals in the input</em>.</p>

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

<pre>
<strong>Input:</strong> intervals = [[1,3],[2,6],[8,10],[15,18]]
<strong>Output:</strong> [[1,6],[8,10],[15,18]]
<strong>Explanation:</strong> Since intervals [1,3] and [2,6] overlap, merge them into [1,6].
</pre>

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

<pre>
<strong>Input:</strong> intervals = [[1,4],[4,5]]
<strong>Output:</strong> [[1,5]]
<strong>Explanation:</strong> Intervals [1,4] and [4,5] are considered overlapping.
</pre>

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

<ul>
	<li><code>1 &lt;= intervals.length &lt;= 10<sup>4</sup></code></li>
	<li><code>intervals[i].length == 2</code></li>
	<li><code>0 &lt;= start<sub>i</sub> &lt;= end<sub>i</sub> &lt;= 10<sup>4</sup></code></li>
</ul>
Array; Sorting
/**
 * @param {number[][]} intervals
 * @return {number[][]}
 */
var merge = function (intervals) {
    intervals.sort((a, b) => a[0] - b[0]);
    const result = [];
    const n = intervals.length;
    let i = 0;
    while (i < n) {
        const left = intervals[i][0];
        let right = intervals[i][1];
        while (true) {
            i++;
            if (i < n && right >= intervals[i][0]) {
                right = Math.max(right, intervals[i][1]);
            } else {
                result.push([left, right]);
                break;
            }
        }
    }
    return result;
};
