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].

/**
 * 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) { List
    
      list = new ArrayList
     
      (); if(intervals.size()==0){ list.add(newInterval); return list; } int start = newInterval.start; int end = newInterval.end; int indexStart=-1;//需要删除节点开始index int insexEnd=-1;//需要删除节点结束index int nstart = -1;//新建节点 int nend = -1; for(int i=0;i
      
       intervals.get(intervals.size()-1).end){ list = intervals; list.add(newInterval); }else if(end