java 22 lines · 7 steps

Merging overlapping intervals in Java

Sort intervals by start, then sweep once, extending or opening ranges as you go.

Explained by highlit
Intermediate intervals sorting greedy arrays
1public List<int[]> merge(int[][] intervals) {
2 if (intervals.length == 0) {
3 return new ArrayList<>();
4 }
5 
6 Arrays.sort(intervals, Comparator.comparingInt(interval -> interval[0]));
7 
8 List<int[]> merged = new ArrayList<>();
9 int[] current = intervals[0].clone();
10 merged.add(current);
11 
12 for (int[] interval : intervals) {
13 if (interval[0] <= current[1]) {
14 current[1] = Math.max(current[1], interval[1]);
15 } else {
16 current = interval.clone();
17 merged.add(current);
18 }
19 }
20 
21 return merged;
22}
01 / 01
STEP 01

Walkthrough

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Three takeaways
  1. 1Sorting by start makes overlaps detectable with a single left-to-right pass.
  2. 2Overlap means the next start is within the current range's end, so extend rather than append.
  3. 3Cloning the interval you add lets you mutate the running range without corrupting the input.

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