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

这道题可以在Merge Intervals基础上完成,时间复杂度有点高。

 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 List<Interval> insert(List<Interval> intervals, Interval newInterval) {
12         int low = newInterval.start;
13         int high = newInterval.end;
14         ListIterator<Interval> iterator = intervals.listIterator();
15         
16         while(iterator.hasNext()){
17             Interval interval = iterator.next();
18             if(high < interval.start){
19                 iterator.previous();
20                 iterator.add(new Interval(low, high));
21                 return intervals;
22             }
23             if(low > interval.end)
24                 continue;
25             else{
26                 low = Math.min(low, interval.start);
27                 high = Math.max(high, interval.end);
28                 iterator.remove();
29             }
30         }//while
31         intervals.add(new Interval(low, high));
32         
33         return intervals;
34     }    
35 }

 

posted on 2015-01-28 10:47  luckygxf  阅读(183)  评论(0编辑  收藏  举报

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