[leetcode]729. My Calendar I

Description

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 and do not add the event to the calendar.

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

分析

题目的意思是：实现一个日历，然后添加行程，要求行程的时间不能冲突，即同一时间段只能有一个行程。
方法也很简单，利用set集合，每次插入一个行程的起始时间和终止时间组成的对，但是在每次插入之前需要判断是否与其他行有冲突就行了。

代码

class MyCalendar {
private:
    set> s1;
public:
    MyCalendar() {
        
    }
    
    bool book(int start, int end) {
        for(auto a:s1){
            if(start>=a.second||end<=a.first){
                continue;
            }else{
                return false;
            }
        }
        s1.insert({start,end});
        return true;
    }
};

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