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.

A 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.

Example 1:

MyCalendarThree();MyCalendarThree.book(10, 20); // returns 1MyCalendarThree.book(50, 60); // returns 1MyCalendarThree.book(10, 40); // returns 2MyCalendarThree.book(5, 15); // returns 3MyCalendarThree.book(5, 10); // returns 3MyCalendarThree.book(25, 55); // returns 3Explanation: 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 becauseeg. [10, 20), [10, 40), and [5, 15) are still triple booked. 

Note:

• 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，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： 

public:    MyCalendarThree() {}        int book(int start, int end) {        ++freq[start];        --freq[end];        int cnt = 0, mx = 0;        for (auto f : freq) {            cnt += f.second;            mx = max(mx, cnt);        }        return mx;    }    private:    map
      
        freq;};