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.

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

Example 1:

MyCalendarThree();
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
Explanation: 
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.

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].

思路:本题是要求出当前并发的日期区间的最大值,即下图中最多有多少个重叠段.
例如:第五次操作调用MyCalendarThree.book(5, 10);
[5,10)与[5,15)是两个重叠段,然而为何返回3呢?往上看,[5,15),[10,40)与[10,20)是三个重叠段.
我们可以在timeline中记录每个事件的start和end,每个start表示在这个时间点增加了一个事件(++),每个end表示在此刻结束了一个事件(--).将他们记录在map中,key为时间点,value为事件数.
这里用到map的重要特性:map中记录是按key值排序的.
因此求多少段重叠,只需遍历map,将每个时刻的事件数求和,最后记录下这个过程中事件数最大值是多少即可.

IMG_7170.jpg

class MyCalendarThree {
public:
    int book(int start, int end) {
        timeline[start]++;  //一个新事件在[start]处开始(首次调用 operator[] 以零初始化value域)
        timeline[end]--;  //一个新事件在[end]处结束
        int max = 0, ongoing = 0;
        for (pair<int, int> t : timeline) {
            ongoing += t.second;  //记录正在进行的事件数
            if (ongoing > max) max = ongoing;  //记录事件数最大值
        }
        return max;
    }
private:
    map<int, int> timeline;
};