leetcode-253.会议室II

题目描述

思路

参考：区间处理-会议室 II（python）_给定一个闭区间列表meetings-优快云博客

最小堆minHeap，先按start排序，然后维护一个minHeap，堆顶元素是会议结束时间最早的区间，也就是end最小。每次比较top元素的end时间和当前元素的start时间，如果start>=end，说明该上一个room可以结束接下来被当前会议区间使用。最后返回堆的大小就是所需的房间数

import heapq
class Solution:
    def minMeetingRooms(self, intervals):
        if not intervals:
            return 0
        intervals.sort(key=lambda x: x[0])
        heap = []
        heapq.heappush(heap, intervals[0][1])
        for i in range(1,len(intervals)):
            if intervals[i][0]>=heap[0]:
                heapq.heappop(heap)
            heapq.heappush(heap, intervals[i][1])
        return len(heap)

if __name__ == '__main__':
    s = Solution()
    intervals = [[0,30], [5,10], [15,20]]
    # intervals = [[7,10],[2,4]]
    print(s.minMeetingRooms(intervals))

import heapq
class Solution:
    def minMeetingRooms(self, intervals):
        if not intervals:
            return 0
        intervals.sort(key=lambda x: x[0])
        heap = []
        heapq.heappush(heap, intervals[0][1])
        for inter in intervals[1:]:
            if inter[0] >= heap[0]:
                heapq.heappop(heap)
            heapq.heappush(heap,inter[1])
        return len(heap)

if __name__ == '__main__':
    s = Solution()
    intervals = [[0 ,30], [5 ,10], [15 ,20]]
    # intervals = [[7,10],[2,4]]
    print(s.minMeetingRooms(intervals))