程序媛詹妮弗
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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:

Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
Output: [[1,5],[6,9]]

Example 2:

Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
Output: [[1,2],[3,10],[12,16]]
Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].

 

题意:

给定一堆区间,插入一个新区间。

 

Solution1:Sort

 

code

 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 class Solution {
11     public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
12         List<Interval> result = new ArrayList<Interval>();
13         for (Interval i : intervals) {
14             if (newInterval == null || i.end < newInterval.start)
15                 result.add(i);
16             else if (i.start > newInterval.end) {
17                 result.add(newInterval);
18                 result.add(i);
19                 newInterval = null;
20             } else {
21                 newInterval.start = Math.min(newInterval.start, i.start);
22                 newInterval.end = Math.max(newInterval.end, i.end);
23             }
24         }
25         if (newInterval != null)
26             result.add(newInterval);
27         return result;
28     }
29 }

 

改版了新的signature之后:

 

posted on 2019-05-14 07:52  程序媛詹妮弗  阅读(143)  评论(0编辑  收藏  举报