732. My Calendar III


Implement a MyCalendarThree class to store your events. A new event can always be added.

Your class will have one method, book(int start, int end). Formally, this represents a booking on the half open interval [start, end), the range of real numbers x such that start <= x < end.

K-booking  happens when K events have some non-empty intersection (ie., there is some time that is common to all K events.)

For each call to the method MyCalendar.book, return an integer K representing the largest integer such that there exists a K-booking in the calendar.

Your class will be called like this: MyCalendarThree cal = new MyCalendarThree(); MyCalendarThree.book(start, end)

Example 1:

MyCalendarThree.book(10, 20); // returns 1
MyCalendarThree.book(50, 60); // returns 1
MyCalendarThree.book(10, 40); // returns 2
MyCalendarThree.book(5, 15); // returns 3
MyCalendarThree.book(5, 10); // returns 3
MyCalendarThree.book(25, 55); // returns 3
The first two events can be booked and are disjoint, so the maximum K-booking is a 1-booking.
The third event [10, 40) intersects the first event, and the maximum K-booking is a 2-booking.
The remaining events cause the maximum K-booking to be only a 3-booking.
Note that the last event locally causes a 2-booking, but the answer is still 3 because
eg. [10, 20), [10, 40), and [5, 15) are still triple booked.



  • The number of calls to MyCalendarThree.book per test case will be at most 400.
  • In calls to MyCalendarThree.book(start, end)start and end are integers in the range [0, 10^9].


这道题是之前那两道题My Calendar IIMy Calendar I的拓展,论坛上有人说这题不应该算是Hard类的,但实际上如果没有之前那两道题做铺垫,直接上这道其实还是还蛮有难度的。这道题博主在做完之前那道,再做这道一下子就做出来了,因为用的就是之前那道My Calendar II的解法二,具体的讲解可以参见那道题,反正博主写完那道题再来做这道题就是秒解啊,参见代码如下:


class MyCalendarThree {
    MyCalendarThree() {}
    int book(int start, int end) {
        int cnt = 0, mx = 0;
        for (auto f : freq) {
            cnt += f.second;
            mx = max(mx, cnt);
        return mx;
    map<int, int> freq;



My Calendar II

My Calendar I





LeetCode All in One 题目讲解汇总(持续更新中…)

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