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].

思路：插入合并就可以，相比于上一题。这题还要简单一些。

详细代码例如以下：

/** * 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
        
         ();    	//边界情况        if(intervals.size() == 0){        	list.add(newInterval);        	return list;        }        //循环推断        for(int i = 0; i < intervals.size();i++){        	//假设新的区间结束值小于開始值。则直接插入前面。后面依次插入就可以        	if(newInterval.end < intervals.get(i).start){        		list.add(newInterval);        		for(int j = i; j < intervals.size(); j++){        			list.add(intervals.get(j));        		}        		break;        	}        	//新的区间開始点大于结束点。则当前点直接加入结果集        	else if(newInterval.start > intervals.get(i).end){                list.add(intervals.get(i));            }        	//须要合并的情况        	else{        		//合并区间                newInterval.start = Math.min(newInterval.start,intervals.get(i).start);                newInterval.end = Math.max(newInterval.end,intervals.get(i).end);            }            if(i == intervals.size() - 1){//假设是最后一个数据。也加入结果集中                list.add(newInterval);            }        }        return list;    }}