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].
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/** * 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 class Solution { public List<Interval> insert(List<Interval> intervals, Interval newInterval) { List<Interval> re = new ArrayList<>(); int i = 0; for (; i < intervals.size(); ++i) { Interval temp = intervals.get(i); if (temp.end >= newInterval.start) break; else re.add(temp); } Interval add = new Interval(newInterval.start, newInterval.end); for (; i < intervals.size(); ++i) { Interval temp = intervals.get(i); if (temp.end < newInterval.start || temp.start > newInterval.end) break; else { add.start = Math.min(add.start, temp.start); add.end = Math.max(add.end, temp.end); } } re.add(add); for (; i < intervals.size(); ++i) re.add(intervals.get(i)); return re; } }