LeetCode解题报告（六）-- 732. My Calendar III

Problem

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)

Example1:

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].

Analysis

本题是Calendar系列的第三题，难度比起前两题有了提升，但是本质上是一样的。我在Calendar Ⅱ中提到了一个多级的集合，这道题的任务就是统计多级集合的级数。但是由于区间重叠的次数是未知的，而且，大量创建vector的会使得代码的效率降低，因此我寻求是否有另一种更加高效简便的方法。
答案当然是有的。我们先来考虑一个情景：一个人参加多个视频会议，每次进入会议室之前他必须先输入账号和密码（登陆），每次结束会议后需要输入账号和密码（登出），那么在任意时刻中，我们只需要统计当前账户登陆的次数就可以知道，该人同时在参加多少个会议。
回到这道题，对于整一个时间轴，我们使用一个map来存储，key是事件的start或end，value是该时间点在线的次数。于是，我们就得到了一条时间轴，我们只需要找出在时间轴上最大的在线数量，即是答案了。

Solution

使用map来存储一条时间轴，原因是map具有对key自动排序的功能。对于每一个需要判断的区间，我们首先在时间轴上进行登录（+1）和登出（-1）操作。然后我们遍历这个map，使用一个cursor加上每个时间点对应的value，找出其中最大的在线次数即可。

Code

class MyCalendarThree {
public:
    MyCalendarThree() {

    }

    int book(int start, int end) {
        timeline[start]++;
        timeline[end]--;

        int max = 0;
        int cursor = 0;
        for(auto iter = timeline.begin(); iter != timeline.end(); iter++) {
            cursor += iter->second;
            max = max >= cursor ? max : cursor;
        }
        return max;
    }

private:
    map<int, int> timeline;
};

/**
 * Your MyCalendarThree object will be instantiated and called as such:
 * MyCalendarThree obj = new MyCalendarThree();
 * int param_1 = obj.book(start,end);
 */

运行时间：88ms，超过40.97%的CPP代码。

Summary

Calendar系列的三道题是难度递增的，他们的思路也是一致的。难点还是在第三题中，如何把问题抽象出一个开会的模式，只要理解了这个模式，实现起来就方便很多了。本题的分析就到这里，谢谢！