Leetcode 252, 253. Meeting Rooms

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

# Definition for an interval.
# class Interval(object):
#     def __init__(self, s=0, e=0):
#         self.start = s
#         self.end = e

class Solution(object):
    def canAttendMeetings(self, intervals):
        """
        :type intervals: List[Interval]
        :rtype: bool
        """
        if not intervals:
            return True
        
        intervals.sort(key=lambda x:x.start)
        
        for i, x in enumerate(intervals):
            if i>0 and x.start < intervals[i-1].end:
                return False
        
        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。

# Definition for an interval.
# class Interval(object):
#     def __init__(self, s=0, e=0):
#         self.start = s
#         self.end = e

class Solution(object):
    def minMeetingRooms(self, intervals):
        """
        :type intervals: List[Interval]
        :rtype: int
        """
        if not intervals:
            return 0
            
        intervals.sort(key=lambda x:x.start)
        endTimes = []
        
        for intv in intervals:
            newroom = True
            for i, x in enumerate(endTimes):
                if intv.start >= x:
                    endTimes[i] = intv.end
                    newroom = False
                    break
            
            if newroom == True:
                endTimes.append(intv.end)
        
            
        return len(endTimes)