Given a setNon-overlappingIntervals, 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 [], [], insert and merge [] in as [], [].
Example 2:
Given [], [], [], [], [], [], insert and merge [] in as [], [], [12, 16].
This is because the new interval [] overlaps with [], [], [].
Https://oj.leetcode.com/problems/insert-interval/
Idea 1: Create a New Result to return and traverse the original line segment group (note that it was originally sorted by START). Therefore, check whether the new interval overlaps with each other, add new results (the results must also be sorted by START ).
public class Solution { public ArrayList<Interval> insert(ArrayList<Interval> intervals, Interval newInterval) { if (newInterval == null) return intervals; ArrayList<Interval> res = new ArrayList<Interval>(); for (Interval each : intervals) { if (each.end < newInterval.start) res.add(each); else if (each.start > newInterval.end) { res.add(newInterval); newInterval = each; } else { newInterval = new Interval(Math.min(each.start, newInterval.start), Math.max(each.end, newInterval.end)); } } res.add(newInterval); return res; } public static void main(String[] args) { // Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as Interval one = new Interval(1, 2); Interval two = new Interval(3, 5); Interval three = new Interval(6, 7); Interval four = new Interval(8, 10); Interval five = new Interval(12, 16); ArrayList<Interval> intervals = new ArrayList<Interval>(); intervals.add(one); intervals.add(two); intervals.add(three); intervals.add(four); intervals.add(five); System.out.println(new Solution().insert(intervals, new Interval(4, 9))); }}
Refer:
Http://www.programcreek.com/2012/12/leetcode-insert-interval/
Http://www.cnblogs.com/TenosDoIt/p/3715013.html