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].

题意：该题与56题非常相似，只是题意给出每一个区间的start是递增的。所以我们不须要排序。该题须要我们加入一个区间。然后进行融合。

这题在56题的基础上添加了区间推断的复杂度。

包括例如以下如所看到的的六种情况。

而这六种情况又能够合并成三种解决方案。看例如以下代码：

/** * 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
      
        insert(List
       
         intervals, Interval newInterval) {       // 先推断newInterval是否在intervals的范围内        if (newInterval == null)            return intervals;        int len = intervals.size();        if (len == 0)        {            intervals.add(newInterval);            return intervals;        }        List
        
          res=new ArrayList
         
          ();        for(Interval interval:intervals)        {            if(interval.end
          
           newInterval.end)//newInterval插入最前端的情况            {                res.add(newInterval);                newInterval=interval;//这个地方非常重要。就是找到了待插入区间位置。指定新的newInterval，由于intervals中的区间也可能有相交的地方，须要融合。            }else if(interval.start<=newInterval.end||interval.end>=newInterval.start)//有重合部分的四种情况            {                newInterval=new Interval(Math.min(interval.start,newInterval.start),Math.max(interval.end,newInterval.end));            }        }        res.add(newInterval);        return res;    }}