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> result = new ArrayList<Interval>(); 13 if (intervals.size()==0){ 14 result.add(newInterval); 15 return result; 16 } 17 18 int next = 0; 19 while (next<intervals.size() && intervals.get(next).end < newInterval.start){ 20 result.add(intervals.get(next)); 21 next++; 22 } 23 24 while (next<intervals.size() && intervals.get(next).start <= newInterval.end){ 25 newInterval.start = Math.min(intervals.get(next).start,newInterval.start); 26 newInterval.end = Math.max(intervals.get(next).end,newInterval.end); 27 next++; 28 } 29 30 result.add(newInterval); 31 32 while (next<intervals.size()){ 33 result.add(intervals.get(next)); 34 next++; 35 } 36 37 return result; 38 } 39 }