Description (leetcode)

Given an array of intervals where intervals[i] = [start_i, end_i], merge all overlapping intervals, and return an array of the non-overlapping intervals that cover all the intervals in the input.

Example 1:

Input: intervals = [[1,3],[2,6],[8,10],[15,18]]

Output: [[1,6],[8,10],[15,18]]

Explanation: Since intervals [1,3] and [2,6] overlap, merge them into [1,6].

Example 2:

Input: intervals = [[1,4],[4,5]]

Output: [[1,5]]

Explanation: Intervals [1,4] and [4,5] are considered overlapping.

Example 3:

Input: intervals = [[4,7],[1,4]]

Output: [[1,7]]

Explanation: Intervals [1,4] and [4,7] are considered overlapping.

Constraints:

  • 1 <= intervals.length <= 10^4
  • intervals[i].length == 2
  • 0 <= start_i <= end_i <= 10^4

submission

impl Solution {
    // intuition: sort and merge
    // we iterate through the intervals, try merge them with the last
    // element of the merged result, if they have overlap, merge them.
    // add mut to intervals arg so we can sort them
    pub fn merge(mut intervals: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
        // we must annotate this variable to compile
        let mut ret: Vec<Vec<i32>> = vec![];
        intervals.sort();
        for interval in intervals {
            match ret.last_mut() {
                // if and only if merged vector have last element and it
                // overlaps with the current interval, we can merge them.
                Some(ref mut last) if interval[0] <= last[1] => {
                    last[1] = last[1].max(interval[1]);
                }
                // any other case, we can't merge them
                _ => ret.push(interval),
            }
        }
        ret
    }
}