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

Your class will have the 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 double booking happens when two events have some non-empty intersection (ie., there is some time that is common to both events.)

For each call to the method MyCalendar.book, return true if the event can be added to the calendar successfully without causing a double booking. Otherwise, return false

Example 1:

MyCalendar();
MyCalendar.book(10, 20); // returns true
MyCalendar.book(15, 25); // returns false
MyCalendar.book(20, 30); // returns true
Explanation: 
The first event can be booked.  The second can‘t because time 15 is already booked by another event.
The third event can be booked, as the first event takes every time less than 20, but not including 20.

Note:

• The number of calls to MyCalendar.book per test case will be at most 1000.
• In calls to MyCalendar.book(start, end), start and end are integers in the range [0, 10^9].

日程設計
要求兩個日程範圍不存在重復。
每次添加時需要和以往所有的日程進行比較。才可以確實是否可以添加。
現羅列出重疊的情況。
a為已存在日程，b為待添加日程。
a　　 a0 l------------l a1
b 　　　 b0 l----l b1

a a0 l------------l a1
b 　　b0 l---------l b1

a 　 a0 l------------l a1
b b0 l------l b1

根據以上重疊情況，可以得到一個抽象的描述
max(a0, b0) < min(a1, b1)

根據這個表達式進行判斷即可，

class MyCalendar {
public:
    vector<pair<int, int>> pvec;
public:
    MyCalendar() {
        
    }
    
    bool book(int start, int end) {
        for (auto p : pvec) {
            if (max(p.first, start) < min(p.second, end))
                return false;
        }
        pvec.push_back({start, end});
        return true;
    }
};
// 108 ms
/**
 * Your MyCalendar object will be instantiated and called as such:
 * MyCalendar obj = new MyCalendar();
 * bool param_1 = obj.book(start,end);
 */

[LeetCode] My Calendar I