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]
.
/** * Definition for an interval. * public class Interval { * int start; * int end; * Interval() { start = 0; end = 0; } * Interval(int s, int e) { start = s; end = e; } * } */ public List<Interval> insert(List<Interval> intervals, Interval newInterval) { List<Interval> result = new ArrayList<>(); for(Interval interval:intervals) { if(interval.end < newInterval.start) { result.add(interval); } else if(interval.start > newInterval.end) { result.add(newInterval); newInterval = interval; } else { newInterval = new Interval(Math.min(interval.start, newInterval.start), Math.max(interval.end, newInterval.end)); } } result.add(newInterval); return result; }