LeetCode 737.My Calendar III

题目描述

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 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 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.

思路分析
每次向时间轴里插入一个有向线段，返回此时最多的重合次数
代码

 class MyCalendarThree {TreeMap<Integer,Integer> timeline = new TreeMap<Integer,Integer>();public MyCalendarThree() {}public int book(int start, int end) {timeline.put(start,timeline.containsKey(start) ? timeline.get(start)+1 : 1);timeline.put(end,timeline.containsKey(end) ? timeline.get(end)-1 :-1);int result = 0,cur = 0;for(int i:timeline.values()){result = Math.max(result,cur=cur+i);}return result;}
}

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