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
impl Solution {
    pub fn merge(mut intervals: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
        intervals.sort_unstable_by(|a, b| a[0].cmp(&b[0]));
        let n = intervals.len();
        let mut res = vec![];
        let mut i = 0;
        while i < n {
            let l = intervals[i][0];
            let mut r = intervals[i][1];
            i += 1;
            while i < n && r >= intervals[i][0] {
                r = r.max(intervals[i][1]);
                i += 1;
            }
            res.push(vec![l, r]);
        }
        res
    }
}
