253. Meeting Room II

https://leetcode.com/problems/meeting-rooms-ii/

solution

• heap

# 每个会议入堆前都判断堆里是否有会议结束，如果有就让它出堆。

class Solution:
    def min_meeting_rooms(self, intervals: List[Interval]) -> int:
        if not intervals:
            return 0

        free_rooms = []
        intervals.sort(key= lambda x: x.start)
        heapq.heappush(free_rooms, intervals[0].end)  # 入堆的是结束时间, 便于判断是否已结束

        for i in intervals[1:]:
            # 注意是if, 不是while. 如果没有overlap, 则可以使用原本的会议室. pop相当于原本占据会议室的人离开
            if free_rooms[0] <= i.start:
                heapq.heappop(free_rooms)

            # 新的会议室加入堆
            heapq.heappush(free_rooms, i.end)

        return len(free_rooms)

时间复杂度：O() 空间复杂度：O()

• 扫描线

# 一个计数器cnt表示当前正在召开的会议数量，然后从小到大遍历所有的时间点。若当前时间点有会议召开，那么就将cnt加上1，反之，若当前时间有会议结束，那么就将cnt减去1

class Solution:
    def min_meeting_rooms(self, intervals: List[Interval]) -> int:
        res = 0
        count = 0
        time = []
        # 更直接方法是开始时间+1, 结束时间-1
        for interval in intervals:
            time.append([interval.start, 1])
            time.append([interval.end, -1])

        time.sort(key=lambda x: x[0])
        for x, status in time:
            count += status
            res = max(res, count)
        return res

时间复杂度：O() 空间复杂度：O()

• 前缀和

class Solution:
    def minMeetingRooms(self, intervals):
        room = {}
        # 开始时间+1，结束时间-1
        for i in intervals:
            room[i.start] = room.get(i.start, 0) + 1
            room[i.end] = room.get(i.end, 0) - 1

        ans = 0
        tmp = 0
        for i in sorted(room.keys()):
            tmp = tmp + room[i]
            ans = max(ans, tmp)
        return ans

follow up

*252. Meeting Rooms