My Calendar II问题及解法

问题描述：

Implement a MyCalendarTwo class to store your events. A new event can be added if adding the event will not cause a triple booking.

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 triple booking happens when three events have some non-empty intersection (ie., there is some time that is common to all 3 events.)

For each call to the method MyCalendar.book, return true if the event can be added to the calendar successfully without causing a triplebooking. Otherwise, return false and do not add the event to the calendar.

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

示例:

MyCalendar();
MyCalendar.book(10, 20); // returns true
MyCalendar.book(50, 60); // returns true
MyCalendar.book(10, 40); // returns true
MyCalendar.book(5, 15); // returns false
MyCalendar.book(5, 10); // returns true
MyCalendar.book(25, 55); // returns true
Explanation: 
The first two events can be booked.  The third event can be double booked.
The fourth event (5, 15) can't be booked, because it would result in a triple booking.
The fifth event (5, 10) can be booked, as it does not use time 10 which is already double booked.
The sixth event (25, 55) can be booked, as the time in [25, 40) will be double booked with the third event;
the time [40, 50) will be single booked, and the time [50, 55) will be double booked with the second event.

问题分析：

本题与My Calendar I 类似，只不过我先保存两次预定相交的部分，然后再判断新预定的内容是否不符合要求。

过程详见代码：

class MyCalendar {
	vector> books;
public:
	bool book(int start, int end) {
		for (pair p : books)
		if (max(p.first, start) < min(end, p.second)) return false;
		books.push_back({ start, end });
		return true;
	}
};

class MyCalendarTwo {
public:
	vector> books;
	MyCalendarTwo() {

	}

	bool book(int start, int end) {
		MyCalendar overlaps;
		for (pair p : books) {
			if (max(p.first, start) < min(end, p.second)) { // overlap exist
				pair overlapped = getOverlap(p.first, p.second, start, end);
				if (!overlaps.book(overlapped.first, overlapped.second)) return false; // overlaps overlapped
			}
		}
		books.push_back({ start, end });
		return true;
	}
	pair getOverlap(int s0, int e0, int s1, int e1) {
		return{ max(s0, s1), min(e0, e1) };
	}
};