57. 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]

 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<Interval>();
13         int len = intervals.size();
14         int i=0;
15         while(i<len&&intervals.get(i).end<newInterval.start){
16             res.add(intervals.get(i++));
17         }
18         while(i<len&&intervals.get(i).start<=newInterval.end){
19             newInterval.start = Math.min(newInterval.start,intervals.get(i).start);
20             newInterval.end = Math.max(newInterval.end,intervals.get(i).end);
21             i++;
22         }
23         res.add(newInterval);
24         while(i<len){
25             res.add(intervals.get(i++));
26         }
27         return res;
28     }
29 }
30 //the run time complexity could be O(n),the space complexity could be O(n);