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LeetCode 731 My Calendar II

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LeetCode



You are implementing a program to use as your calendar. We can add a new event if adding the event will not cause a triple booking.

A triple booking happens when three events have some non-empty intersection (i.e., some moment is common to all the three events.).

The event can be represented as a pair of integers start and end that represents a booking on the half-open interval [start, end), the range of real numbers x such that start <= x < end.

Implement the MyCalendarTwo class:

MyCalendarTwo() Initializes the calendar object. boolean book(int start, int end) Returns true if the event can be added to the calendar successfully without causing a triple booking. Otherwise, return false and do not add the event to the calendar.

Example 1:

Input
[“MyCalendarTwo”, “book”, “book”, “book”, “book”, “book”, “book”]
[[], [10, 20], [50, 60], [10, 40], [5, 15], [5, 10], [25, 55]]
Output
[null, true, true, true, false, true, true]

Explanation
MyCalendarTwo myCalendarTwo = new MyCalendarTwo();
myCalendarTwo.book(10, 20); // return True, The event can be booked.
myCalendarTwo.book(50, 60); // return True, The event can be booked.
myCalendarTwo.book(10, 40); // return True, The event can be double booked.
myCalendarTwo.book(5, 15);  // return False, The event ca not be booked, because it would result in a triple booking.
myCalendarTwo.book(5, 10); // return True, The event can be booked, as it does not use time 10 which is already double booked.
myCalendarTwo.book(25, 55); // return True, The event can be booked, as the time in [25, 40) will be double booked with the third event, the time [40, 50) will be single booked, and the time [50, 55) will be double booked with the second event.


Constraints:

0 <= start < end <= 10<sup>9</sup> At most 1000 calls will be made to book.

暴力法算法思路(推荐)：

存储所有区间和重合区间

注意事项：

  1. 区间比较难写，用符合加入会议条件的相反来写，not (start >= root.end or end <= root.start)

Python代码：

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class MyCalendar(TestCases):
    def __init__(self):
        self.calendar = []
        self.overlaps = []

    def book(self, start: int, end: int) -> bool:
        for s, e in self.overlaps:
            if not (start >= e or end <= s):
                return False

        for s, e in self.calendar:
            if not (start >= e or end <= s):
                self.overlaps.append((max(start, s), min(end, e)))
        self.calendar.append((start, end))
        return True

算法分析：

时间复杂度为O(n2), 空间复杂度O(n)

算法II解题思路：

有重复区间，考虑用meeting rooms的方法二。用一个map来存储endpoint包括start和end的频率，若遇到start，map[start]++, 若遇到end, map[end]–，插入一个区间后，遍历所有endpoints，若超过3就返回False
虽然复杂度稍差，系数更大。但此法更有推广性，如果允许重复会议更多，此法可扩展

注意事项：

  1. 先插入有序区间，然后统计看是否有重复区间超过2.
  2. 若是False，用remove这个函数要删除刚插入的。

Python代码：

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class MyCalendar(TestCases):
    def __init__(self):
        self.calendar = []

    def book(self, start: int, end: int) -> bool:
        bisect.insort(self.calendar, (start, 1))
        bisect.insort(self.calendar, (end, -1))
        active_meeting = 0
        for endpoint, _type in self.calendar:
            if _type == 1:
                active_meeting += 1
            else:
                active_meeting -= 1
            if active_meeting >= 3:
                self.calendar.remove((start, 1))
                self.calendar.remove((end, -1))
                return False
        return True

算法分析：

时间复杂度为O(n2), 空间复杂度O(n)

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