zoukankan      html  css  js  c++  java
  • LeetCode 731. My Calendar II

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

    题目:

    Implement a MyCalendarTwo class to store your events. A new event can be added if adding the event will not cause a triple booking.

    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.

    triple booking happens when three events have some non-empty intersection (ie., there is some time that is common to all 3 events.)

    For each call to the method MyCalendar.book, return true if the event can be added to the calendar successfully without causing a triple 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(50, 60); // returns true
    MyCalendar.book(10, 40); // returns true
    MyCalendar.book(5, 15); // returns false
    MyCalendar.book(5, 10); // returns true
    MyCalendar.book(25, 55); // returns true
    Explanation: 
    The first two events can be booked.  The third event can be double booked.
    The fourth event (5, 15) can't be booked, because it would result in a triple booking.
    The fifth event (5, 10) can be booked, as it does not use time 10 which is already double booked.
    The sixth event (25, 55) 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.

    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].

    题解:

    Maintian the ongoing event. For each event, at start, mark this point ongoing events count +1. At end, mark this point ongoing events count -1.

    Every time when booking, first add this event, and iterate all the ongoing events, accumlating the current ongoing event counts.

    If the count is larger than 2, that means there are at least 3 events overlapped. remove this event and return false.

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

    Space: O(n).

    AC Java:

     1 class MyCalendarTwo {
     2     TreeMap<Integer, Integer> tm;
     3     
     4     public MyCalendarTwo() {
     5         tm = new TreeMap<>();    
     6     }
     7     
     8     public boolean book(int start, int end) {
     9         tm.put(start, tm.getOrDefault(start, 0)+1);
    10         tm.put(end, tm.getOrDefault(end, 0)-1);
    11         int ongoing = 0;
    12         for(int count : tm.values()){
    13             ongoing += count;
    14             if(ongoing > 2){
    15                 tm.put(start, tm.get(start)-1);
    16                 if(tm.get(start) == 0){
    17                     tm.remove(start);
    18                 }
    19                 
    20                 tm.put(end, tm.get(end)+1);
    21                 if(tm.get(end) == 0){
    22                     tm.remove(end);
    23                 }
    24                 
    25                 return false;
    26             }
    27         }
    28         
    29         return true;
    30     }
    31 }
    32 
    33 /**
    34  * Your MyCalendarTwo object will be instantiated and called as such:
    35  * MyCalendarTwo obj = new MyCalendarTwo();
    36  * boolean param_1 = obj.book(start,end);
    37  */

    类似My Calendar IMy Calendar III.

  • 相关阅读:
    IDEA的Debug详解
    websocket学习(转载)
    Shiro授权及注解式开发
    Redis分布式缓存安装和使用
    JEESZ-SSO解决方案
    英语是学习Java编程的基础吗
    深入分析 ThreadLocal 内存泄漏问题
    这些JVM命令配置参数你知道吗?
    安全开发Java动态代理
    学java编程软件开发,非计算机专业是否能学
  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/11875122.html
Copyright © 2011-2022 走看看