Question

我可能会比这更难......但我的任务是创建一个实现Comparable的showtimes类。有了这个，它构建了几个不同剧院的列表，其中包括时间等等。然后我必须使用compareTo（）方法对其进行排序，首先是它所显示的电影，以及电影。标题。我有以下代码......当我尝试输出排序结果时出现混乱。我把它输出到比较字符串的地方输出 - 或+数字。

public class ShowTimeTest {


public static void main(String[] args) {
    ShowTime[] movieTimes= new ShowTime[20];
    // TODO Auto-generated method stub
    //******************************************************
    //Instantiates 20 showtime objects
    //******************************************************



ShowTime cinemark1 = new ShowTime("Cinemark", "Lorde", "Friday @ 3:30", "Saturday @ 7:00", "Sunday @ 6:00","R", 1, 2);
ShowTime cinemark2 = new ShowTime("Amstar Cinemas", "Star Wars", "Thursday @ 12:00AM", "Monday @ 4:00PM", "Sunday @ 3:45PM","R", 1, 2);
ShowTime cinemark3 = new ShowTime("Fayette Movies", "Pokemon", "Friday @ 3:30", "Saturday @ 7:00", "Sunday @ 6:00","R", 1, 2);
ShowTime cinemark4 = new ShowTime("Dollar Theatre", "Reincarnated", "Friday @ 3:30", "Saturday @ 7:00", "Sunday @ 6:00","R", 1, 2);
ShowTime cinemark5 = new ShowTime("Rad Chads Cinemas", "Lorde", "Friday @        3:30",     "Saturday @ 7:00", "Sunday @ 6:00","R", 1, 2);
movieTimes[0] = cinemark1;
movieTimes[1] = cinemark2;
movieTimes[2] = cinemark3;
movieTimes[3] = cinemark4;
movieTimes[4] = cinemark5;

for(int i = 1; i < 5; i ++){
    System.out.println(movieTimes[i].compareTo(movieTimes[i-1]));

 }
}
}

续...

public class ShowTime implements Comparable<ShowTime>{

public String name, title, rating, showTime1, showTime2, showTime3;
public double ticket, adultTicket;
public String[] times = new String[3];

public ShowTime(String theatreName, String movieTitle, String showTime1, String showTime2, String showTime3 , String movieRating, double childTicket, double aTicket)
{
    // TODO Auto-generated constructor stub
name = theatreName; 
title = movieTitle;
times[0] = showTime1;
times[1] = showTime2;
times[2] = showTime3;
rating = movieRating;
ticket = childTicket;
adultTicket = aTicket;
}
public String toString(){
    String out = "Theatre: ";
    out += name+ "  " + "Movie: " + title + " " + "Rating " + rating + " " + "ShowTimes " + times[0] + " " + times[1] + " " +  times[2]
            +  " " + "Adult Cost: " + adultTicket + " " + "Child Ticket: " +ticket;
    return out;
}
public int compareTo(ShowTime other) {
    // TODO Auto-generated method stub
    //sorts the showtimes by theater(NAME) and movie(TITLE)
    int result;

    if(name.equals(((ShowTime)other).name)){
        result = title.compareTo(((ShowTime)other).title);
    }

    else{
        result = name.compareTo(((ShowTime)other).name);
    }
    return result;
}
}

Answer 1

使您的类实现Comparable接口的重点是，您可以对ShowTime的数组进行排序。你已经完成了所有艰苦的工作。现在只需申请Arrays.sort，坐下来欣赏电影。

由于这是一个教育节目，让我100％透明地说：你已经拥有一个Object[]，其元素实现Comparable，即movieTimes。所以，你需要做的就是添加单行：

Arrays.sort(movieTimes);

请注意，在您为影院填充20部电影（或将其声明为仅包含5部电影的迷你电影）之前，这将失败。

Answer 2

您可以使用Collections类：

List<ShowTime> list = Arrays.asList(array);
Collections.sort(list);
movieTimes = list.toArray();

使用compareTo（）对对象数组进行排序

2 个答案: