Leetcode 252, 253. Meeting Rooms

Posted 2020-09-18 lettuan

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

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

 

所有meeting time按照start排序之后，逐个检查当前的start是不是小于上一个end。如果是的话输出False

 1 # Definition for an interval.
 2 # class Interval(object):
 3 #     def __init__(self, s=0, e=0):
 4 #         self.start = s
 5 #         self.end = e
 6 
 7 class Solution(object):
 8     def canAttendMeetings(self, intervals):
 9         """
10         :type intervals: List[Interval]
11         :rtype: bool
12         """
13         if not intervals:
14             return True
15         
16         intervals.sort(key=lambda x:x.start)
17         
18         for i, x in enumerate(intervals):
19             if i>0 and x.start < intervals[i-1].end:
20                 return False
21         
22         return True

第二题，要求需要多少间会议室才能满足要求。考虑real life中安排会议的方法。先按会议的开始时间排序。然后逐个安排。如果当年要安排的会议的开始时间在某个会议的结束时间之后，那么就可以用那一间会议室，而不用另开房间。如果当前要安排的会议的开始时间比所有房间的结束时间都早，那么需要另开一个新房间。注意在安排的新的会议之后，要更新这个房间的结束时间。注意L25的break，这说明如果已经为当前要安排的会议找到了一间，那么就不需要再看其他的房间了。

 1 # Definition for an interval.
 2 # class Interval(object):
 3 #     def __init__(self, s=0, e=0):
 4 #         self.start = s
 5 #         self.end = e
 6 
 7 class Solution(object):
 8     def minMeetingRooms(self, intervals):
 9         """
10         :type intervals: List[Interval]
11         :rtype: int
12         """
13         if not intervals:
14             return 0
15             
16         intervals.sort(key=lambda x:x.start)
17         endTimes = []
18         
19         for intv in intervals:
20             newroom = True
21             for i, x in enumerate(endTimes):
22                 if intv.start >= x:
23                     endTimes[i] = intv.end
24                     newroom = False
25                     break
26             
27             if newroom == True:
28                 endTimes.append(intv.end)
29         
30             
31         return len(endTimes)