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].

[解题思路]

新建一个ArrayList来保存得到的结果。因为对原来的集合进行操作时如下情况需要额外的操作：newInterval插入当前interval之前

1.newInterval.end < curInterval.start, 插入

2.newInterval.end > curInterval.start, continue

3.newInterval 与 curInterval 重合,合并得到新的newInterval 

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 static ArrayList
     
       insert(ArrayList
      
        intervals,12             Interval newInterval) {13         ArrayList
       
         result = new ArrayList
        
         ();14         for (int i = 0; i < intervals.size(); i++) {15             Interval tmp = intervals.get(i);16             if (newInterval.end < tmp.start) {17                 result.add(newInterval);18                 for(int j = i; j < intervals.size(); j++){19                     result.add(intervals.get(j));20                 }21                 return result;22             } else if (newInterval.start > tmp.end) {23                 result.add(tmp);24                 continue;25             } else {26                 newInterval.start = Math.min(tmp.start, newInterval.start);27                 newInterval.end = Math.max(tmp.end, newInterval.end);28             }29         }30         result.add(newInterval);31         return result;32     }33 }