[leetcode]252. Meeting Rooms会议室有冲突吗

Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...] (si < ei), determine if a person could attend all meetings.

Example 1:

Input: [[0,30],[5,10],[15,20]]
Output: false

Example 2:

Input: [[7,10],[2,4]]
Output: true

题目

给定一些区间,判断是否有重合。

还是挺实际的场景,经常在图书馆预定study room的系统,内部也应该是这个逻辑。 当有overlapping的时候,就会报错。

 

思路

任何一组intervals, 若当前start < 之前end,即出现了overlapping 

代码

 1 class Solution {
 2     public boolean canAttendMeetings(Interval[] intervals) {
 3         if(intervals == null || intervals.length==0) return true;
 4         int []start = new int[intervals.length];
 5         int []end = new int[intervals.length];
 6         for(int i = 0; i<intervals.length;i++){
 7             start[i] = intervals[i].start;
 8             end[i] = intervals[i].end;
 9         }
10         Arrays.sort(start);
11         Arrays.sort(end);
12         /* 任何一组intervals, 若当前start < 之前end,即出现了overlapping 
13             | i-1 |
14                 |  i  |   
15         */ 
16         for(int i = 1; i< start.length; i++){
17             if(start[i]<end[i-1]) return false;
18         }
19         return true;
20     }
21 }
原文地址:https://www.cnblogs.com/liuliu5151/p/9823970.html