Meeting Rooms I/II
要点:这题和skyline类似,利用了interval start有序的特点,从左向右处理,用一个heap来动态表示当前占用rooms的时间段,所以heap的size就是room数。具体来说,
- heap是end time的min heap
- 当前?就是和新interval同时使用room的情况
- 如果min end<=新的interval.start,那么同一房间可以被这个interval重用。同时所有heap中end小的都要pop
- 如果min end>新的interval.start<min end:因为新interval.start比所有当前在heap中start的都晚,所以一定与所有heap中interval在某一点都有交集。所有max需要新房间,要push heap
- 所以在heap中的intervals必然都是在某一点互相重叠的
- 当然用endPoint sort + count的方法也可以。都是O(nlgn),但不需要heap
https://repl.it/CfU6/4 (II)
错误点:不管是否pop,都要push新的:either 占新room or replace旧room
https://repl.it/CoP7 (I)
EDIT:
还有一类类似的题,是room (or 资源)有限,如何选最多的meeting。
# Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...] (si < ei), find the minimum number of conference rooms required.
# For example,
# Given [[0, 30],[5, 10],[15, 20]],
# return 2.
# Hide Company Tags Google Facebook
# Hide Tags Heap Greedy Sort
# Hide Similar Problems (H) Merge Intervals (E) Meeting Rooms
from heapq import heappush, heappop
# 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
"""
meetings = []
intervals.sort(key=lambda interval:interval.start)
count = 0
while i in intervals:
if not meetings or meetings[0]>i.start:
heappush(meetings, i.end)
i+=1
else:
heappop(meetings)
count = max(count, len(meetings))
return count
# 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.
# Hide Company Tags Facebook
# Hide Tags Sort
# Hide Similar Problems (H) Merge Intervals (M) Meeting Rooms II
# 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
"""
# intervals = sorted(intervals, key=lambda interval: interval.start)
intervals.sort(key=lambda interval: interval.start)
for i in xrange(1, len(intervals)):
if intervals[i].start<intervals[i-1].end:
return False
return True