LeetCode Meeting Rooms

Posted 2020-06-14

原题链接在这里：https://leetcode.com/problems/meeting-rooms/

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.

For example,
Given [[0, 30],[5, 10],[15, 20]],
return false.

对array进行排序，从i = 2开始判断start是否在前一个end之前, 若是就return false. 完成loop返回true.

Time Complexity: O(nlogn). Space: O(1).

AC Java:

 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 boolean canAttendMeetings(Interval[] intervals) {
12         if(intervals == null || intervals.length == 0){
13             return true;
14         }
15         Arrays.sort(intervals, new Comparator<Interval>(){
16             public int compare(Interval i1, Interval i2){
17                 if(i1.start == i2.start){
18                     return i1.end - i2.end;
19                 }
20                 return i1.start - i2.start;
21             }
22         });
23         
24         for(int i = 1; i<intervals.length; i++){
25             if(intervals[i].start < intervals[i-1].end){
26                 return false;
27             }
28         }
29         return true;
30     }
31 }

 跟上Meeting Rooms II