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<Interval> insert(List<Interval> intervals, Interval newInterval) {

    	 List<Interval> list = new ArrayList<Interval>();

    	//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;

    }

}