Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).
You may assume that the intervals were initially sorted according to their start times.
Example 1:
Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].
Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].
This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].
Solution:
public List<Interval> insert(List<Interval> intervals, Interval newInterval) { intervals.add(newInterval); intervals.sort((i1, i2) -> Integer.compare(i1.start, i2.start)); List<Interval> result = new ArrayList<>(); int start = intervals.get(0).start; int end = intervals.get(0).end; for (Interval i : intervals) { if (i.start <= end) { end = Math.max(i.end, end); } else { result.add(new Interval(start, end)); start = i.start; end = i.end; } } result.add(new Interval(start, end)); return result; }Solution2:
better
public List<Interval> insert(List<Interval> intervals, Interval newInterval) { int i = 0; int len = intervals.size(); while (i < len && intervals.get(i).end < newInterval.start) { i++; } while (i < intervals.size() && intervals.get(i).start <= newInterval.end) { newInterval.start = Math.min(newInterval.start, intervals.get(i).start); newInterval.end = Math.max(newInterval.end, intervals.get(i).end); intervals.remove(i); } intervals.add(i, newInterval); return intervals; }