zoukankan      html  css  js  c++  java
  • 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.

  • 相关阅读:
    Bug测试报告--在线考试系统--金州勇士
    Bug测试报告--食物链教学工具--奋斗吧兄弟
    Jquery对象和dom对象获取html的方法
    mysql中常见的存储引擎和索引类型
    转:spring MVC HTTP406 Not Acceptable
    Mybatis动态建表
    ssm框架插入mysql数据库中文乱码问题解决
    Maven环境下Poi的使用
    【转】Mybatis传多个参数(三种解决方案)
    【译文】用Spring Cloud和Docker搭建微服务平台
  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/11875126.html
Copyright © 2011-2022 走看看