[leetcode] 253. Meeting Rooms II @ python

原题

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.

Example 1:

Input: [[0, 30],[5, 10],[15, 20]]
Output: 2
Example 2:

Input: [[7,10],[2,4]]
Output: 1

解法

最小堆算法. 利用heap[0]永远是队列中最小值这一性质, 我们构建一个heap队列, 遍历intervals, 如果当前的intervals的开始时间在heap[0]之后, 那么两个会议可以用同一个房间, 我们将当前会议的结束时间来替代heap[0]; 否则需要另外开一个房间, 将会议的结束时间加到heap队列里, 最后返回heap的长度.

Time: O(n)
Space: O(1)

代码

# 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
        """              
        intervals.sort(key = lambda x: x.start)
        heap = []
        for i in intervals:
            if heap and i.start >= heap[0]:
                # two meetings can use the same room
                heapq.heapreplace(heap, i.end)
            else:
                # a new room is allocated
                heapq.heappush(heap, i.end)
                
        return len(heap)