Intervals: merge overlapping ranges

You're given a list of intervals like [[1,3],[2,6],[8,10],[15,18]] and asked to merge every pair that overlaps (LeetCode #56). Here [1,3] and [2,6] touch, so they collapse into [1,6]; [8,10] and [15,18] stand alone. The result is [[1,6],[8,10],[15,18]].

The unlock is to sort the intervals by their start. Once they're sorted, overlaps can only happen between an interval and the one most recently merged — you never have to look backwards. Walk the sorted list keeping a "current" merged interval; if the next one starts at or before the current end, stretch the end, otherwise close out the current interval and start a new one.

intervals.sort(key=lambda iv: iv[0])
merged = [intervals[0]]
for start, end in intervals[1:]:
    if start <= merged[-1][1]:      # overlaps the last merged interval
        merged[-1][1] = max(merged[-1][1], end)
    else:
        merged.append([start, end])

Note that touching counts as overlapping: [1,4] and [4,5] merge into [1,5] because 4 <= 4.

Your task

Write merge(intervals) returning a list of [start, end] lists with all overlaps merged, sorted by start. Return plain lists (not tuples). An empty input gives an empty list.

merge([[1,3],[2,6],[8,10],[15,18]])   # [[1, 6], [8, 10], [15, 18]]
merge([[1,4],[4,5]])                   # [[1, 5]]   (touching merges)
merge([[1,4]])                         # [[1, 4]]
merge([])                              # []

Hint: sort first, then sweep once, growing the last merged interval whenever the next one starts at or before its end.