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  • LeetCode 732. My Calendar III

    原题链接在这里:https://leetcode.com/problems/my-calendar-iii/

    题目:

    Implement a MyCalendarThree class to store your events. A new event can always be added.

    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.

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

    For each call to the method MyCalendar.book, return an integer K representing the largest integer such that there exists a K-booking in the calendar.

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

    Example 1:

    MyCalendarThree();
    MyCalendarThree.book(10, 20); // returns 1
    MyCalendarThree.book(50, 60); // returns 1
    MyCalendarThree.book(10, 40); // returns 2
    MyCalendarThree.book(5, 15); // returns 3
    MyCalendarThree.book(5, 10); // returns 3
    MyCalendarThree.book(25, 55); // returns 3
    Explanation: 
    The first two events can be booked and are disjoint, so the maximum K-booking is a 1-booking.
    The third event [10, 40) intersects the first event, and the maximum K-booking is a 2-booking.
    The remaining events cause the maximum K-booking to be only a 3-booking.
    Note that the last event locally causes a 2-booking, but the answer is still 3 because
    eg. [10, 20), [10, 40), and [5, 15) are still triple booked.

    Note:

    • The number of calls to MyCalendarThree.book per test case will be at most 400.
    • In calls to MyCalendarThree.book(start, end)start and end are integers in the range [0, 10^9].

    题解:

    Maintain the ongoing events by plus one at each booking start and minus one at each booling end.

    Accumlate ongoing events count and return the maximum.

    It could array to maintain such events, but input may be sparse. Thus it would be better to use TreeMap.

    Time Complexity: book, O(nlogn). n is size of tm.

    Space: O(n).

    AC Java:

     1 class MyCalendarThree {
     2     TreeMap<Integer, Integer> tm;
     3     
     4     public MyCalendarThree() {
     5         tm = new TreeMap<>();    
     6     }
     7     
     8     public int book(int start, int end) {
     9         tm.put(start, tm.getOrDefault(start, 0)+1);
    10         tm.put(end, tm.getOrDefault(end, 0)-1);
    11         
    12         int res = 0;
    13         int ongoing = 0;
    14         for(int count : tm.values()){
    15             ongoing += count;
    16             res = Math.max(res, ongoing);
    17         }
    18         
    19         return res;
    20     }
    21 }
    22 
    23 /**
    24  * Your MyCalendarThree object will be instantiated and called as such:
    25  * MyCalendarThree obj = new MyCalendarThree();
    26  * int param_1 = obj.book(start,end);
    27  */

    类似My Calendar IMy Calendar II.

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  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/11875126.html
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