731. My Calendar II
Description
You are implementing a program to use as your calendar. We can add a new event if adding the event will not cause a triple booking.
A triple booking happens when three events have some non-empty intersection (i.e., some moment is common to all the three events.).
The event can be represented as a pair of integers startTime
and endTime
that represents a booking on the half-open interval [startTime, endTime)
, the range of real numbers x
such that startTime <= x < endTime
.
Implement the MyCalendarTwo
class:
MyCalendarTwo()
Initializes the calendar object.boolean book(int startTime, int endTime)
Returnstrue
if the event can be added to the calendar successfully without causing a triple booking. Otherwise, returnfalse
and do not add the event to the calendar.
Example 1:
Input ["MyCalendarTwo", "book", "book", "book", "book", "book", "book"] [[], [10, 20], [50, 60], [10, 40], [5, 15], [5, 10], [25, 55]] Output [null, true, true, true, false, true, true] Explanation MyCalendarTwo myCalendarTwo = new MyCalendarTwo(); myCalendarTwo.book(10, 20); // return True, The event can be booked. myCalendarTwo.book(50, 60); // return True, The event can be booked. myCalendarTwo.book(10, 40); // return True, The event can be double booked. myCalendarTwo.book(5, 15); // return False, The event cannot be booked, because it would result in a triple booking. myCalendarTwo.book(5, 10); // return True, The event can be booked, as it does not use time 10 which is already double booked. myCalendarTwo.book(25, 55); // return True, The event 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.
Constraints:
0 <= start < end <= 109
- At most
1000
calls will be made tobook
.
Solutions
Solution 1
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