LeetCode Meeting Rooms

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

题目：

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.

题解：

对array进行排序，从i = 1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:

 /**
  * Definition for an interval.
  * public class Interval {
  *     int start;
  *     int end;
  *     Interval() { start = 0; end = 0; }
  *     Interval(int s, int e) { start = s; end = e; }
  * }
  */
 public class Solution {
     public boolean canAttendMeetings(Interval[] intervals) {
         if(intervals == null || intervals.length == 0){
             return true;
         }
         Arrays.sort(intervals, new Comparator<Interval>(){
             public int compare(Interval i1, Interval i2){
                 if(i1.start == i2.start){
                     return i1.end - i2.end;
                 }
                 return i1.start - i2.start;
             }
         });
 
         for(int i = 1; i<intervals.length; i++){
             if(intervals[i].start < intervals[i-1].end){
                 return false;
             }
         }
         return true;
     }
 }

跟上Meeting Rooms II