Insert Interval

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:

  1. /**
  2.  * Definition for an interval.
  3.  * public class Interval {
  4.  *     int start;
  5.  *     int end;
  6.  *     Interval() { start = 0; end = 0; }
  7.  *     Interval(int s, int e) { start = s; end = e; }
  8.  * }
  9.  */
  10. public class Solution {
  11.   public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
  12.     List<Interval> res = new ArrayList<>();
  14.     int i = 0, n = intervals.size();
  16.     while (i < n && intervals.get(i).end < newInterval.start) {
  17.       res.add(intervals.get(i++));
  18.     }
  20.     while (i < n && intervals.get(i).start <= newInterval.end) {
  21.       newInterval = new Interval(
  22.         Math.min(newInterval.start, intervals.get(i).start),
  23.         Math.max(newInterval.end, intervals.get(i).end)
  24.       );
  25.       i++;
  26.     }
  28.     res.add(newInterval);
  30.     while (i < n) {
  31.       res.add(intervals.get(i++));
  32.     }
  34.     return res;
  35.   }
  36. }