https://leetcode.com/problems/my-calendar-i/
You are implementing a program to use as your calendar. We can add a new event if adding the event will not cause a double booking. A double booking happens when two events have some non-empty intersection (i.e., some moment is common to both events.). The event can be represented as a pair of integers start and end that represents a booking on the half-open interval [start, end), the range of real numbers x such that start <= x < end. Implement the MyCalendar class:
MyCalendar() Initializes the calendar object.
boolean book(int start, int end) Returns 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.
Example 1: Input [“MyCalendar”, “book”, “book”, “book”] [[], [10, 20], [15, 25], [20, 30]] Output [null, true, false, true] Explanation MyCalendar myCalendar = new MyCalendar(); myCalendar.book(10, 20); // return True myCalendar.book(15, 25); // return False, It can not be booked because time 15 is already booked by another event. myCalendar.book(20, 30); // return True, The event can be booked, as the first event takes every time less than 20, but not including 20.
Constraints:
0 <= start < end <= 109
At most 1000 calls will be made to book.
- code
class MyCalendar {
List<int[]> c;
public MyCalendar() {
c = new ArrayList<>();
}
public boolean book(int start, int end) {
for (int[] event: c){
if (start >= event[1] || end <= event[0]) continue;
else return false;
}
c.add(new int[]{start, end});
return true;
}
}
- code need to use Integer, possibly it’s null value
class MyCalendar {
TreeMap<Integer, Integer> calendar;
MyCalendar() {
calendar = new TreeMap();
}
public boolean book(int start, int end) {
Integer prev = calendar.floorKey(start),
next = calendar.ceilingKey(start);
if ((prev == null || calendar.get(prev) <= start) &&
(next == null || end <= next)) {
calendar.put(start, end);
return true;
}
return false;
}
}