Question

我有3节课。我正在学习接口和接口类TravelCost必须具有公共摘要以及方法类型和名称，以便它通过所有三个类保持一致。这三个班级（AirTravelCost，TrainTravelCost，CarTravelCost）将实施TravelCost。我已完成所有设置和测试工作。但是，我假设的测试页面是您通过arrayList输入搜索的最低成本和最短持续时间。我不知道怎么做，因为我以前从未在ArrayList中做过这个。这是测试类中的示例代码：

import java.util.*;

 public class TestTravelCost
 {
   public static void main(String [] args)
   {
    /*Scanner scn = new Scanner(System.in); //scanner object

    System.out.println("Number of Miles: ");
    double numOfMiles = scn.nextDouble();

    System.out.println("Hotel cost per night: ");
    double cost = scn.nextDouble();

    System.out.println("Description: ");
    String description = scn.nextLine();*/

    TravelCost c = new CarTravelCost(400, 200, "Boston");//instantiate object for car travel
    TravelCost t = new TrainTravelCost(6, 60.0, "Boston"); //instantiate object for train travel
    TravelCost a = new AirTravelCost(224, "20110103", "0743" , "20110103", "1153", "Boston");//instantiate object for air travel

    ArrayList<TravelCost> AL = new ArrayList<TravelCost>();//array list for car travel
    AL.add(c);
    AL.add(t);
    AL.add(a);

    for(TravelCost tc : AL)
    {
        System.out.println(tc.toString());
    }
   }
}

输出：前往波士顿的车程为7.272727272727272725小时，费用为210.0 前往波士顿的火车需要6.0小时，费用为70.0 前往波士顿的航空旅行需要1.0166666666666666，费用为243.48888888888888 //这不是正确的计算方法，我不知道我错在哪里，但假设它与最短的持续时间相同。我想我不擅长数学

这是我用于航空旅行的计算方法

    public double getDuration()
{
    //---DEPARTURE---//
    int Dyear = Integer.parseInt(departureDate.substring(0,3)); //2011
    int Dmonth = Integer.parseInt(departureDate.substring(4,5));//01
    int Dday = Integer.parseInt(departureDate.substring(6,7));//03

    int Dhour = Integer.parseInt(departureTime.substring(0,1));//0743
    int Dminute = Integer.parseInt(departureTime.substring(2,3));//1153
    //---ARRIVAL---//
    int Ayear = Integer.parseInt(arrivalDate.substring(0,3)); //2011
    int Amonth = Integer.parseInt(arrivalDate.substring(4,5));//01
    int Aday = Integer.parseInt(arrivalDate.substring(6,7));//03

    int Ahour = Integer.parseInt(arrivalTime.substring(0,1));//0743
    int Aminute = Integer.parseInt(arrivalTime.substring(2,3));//1153

    GregorianCalendar date = new GregorianCalendar(Dyear, Dmonth, Dday, Dhour, Dminute);//departure date & time
    GregorianCalendar time = new GregorianCalendar(Ayear, Amonth, Aday, Ahour, Aminute);//arrival date & time

    //date = arrivalDate - departureDate;//2011-01-03 - 2011-01-03 = 0
    //time = arrivalTime - departureTime;//0734 - 1153 = 410

    double duration = (Math.abs(date.getTimeInMillis() - time.getTimeInMillis()) / 60000.0) / 60.0;
    return duration;
  `enter code here` }

如何在我的代码中获得此结果？

最低成本：火车前往波士顿需要11.0小时，费用为70.0 最短的持续时间：前往波士顿的航班将需要4.166666666666667小时，费用为234.0

Answer 1

您没有显示TravelCost界面，但要达到您想要的效果，至少应该使用getDuration和getCost方法。

public interface TravelCost {
    ... // what you already have in the interface definition
    public double getDuration();
    public double getCost();
}

有了这个，我会创建一个小的虚拟类，实现基本功能，以便在这些属性上具有可比性：

public DummyTC implements TravelCost {
    private double cost;
    private double duration;

    public DummyTC(double cost, double duration) {
        this.cost = cost;
        this.duration = duration;
    }

    public double getDuration() {
        return duration;
    }

    public double getCost() {
        return cost;
    }

    // and other methods/properties imposed by TravelCost
}

这将允许您找到您要找的内容：

// instantiate 2 DummyTC's with impossibly high cost &durations

TravelCost cheapest = new DummyTC(99999999999.99, 99999999999.99);
TravelCost shortest = new DummyTC(99999999999.99, 99999999999.99);

// iterate over the List

for(TravelCost tc : AL) {
    // if present tc is cheaper than cheapest, swap
    if ( tc.getCost() < cheapest.getCost() ) {
        cheapest = tc;
    }
    // if present tc is shorter than shortest, swap
    if ( tc.getDuration() < shortest.getDuration() ) {
        shortest = tc;
    }
}

// at this point cheapest and shortest will contain the cheapest and shortest 
// ways of transportation, so we print them out.

System.out.println(cheapest.toString());
System.out.println(shortest.toString());

另一件事，您的日期处理代码非常令人费解。看看SimpleDateFormat

Date date = null;
Date time = null;
// create a SimpleDateFormat instance for your time/date format
SimpleDateFormat format = new SimpleDateFormat("yyyyMMdd");
try {
    // parse it
    date = format.parse(departureDate);
    // done
} catch (ParseException e) {
    // departureDate could not be parsed, you should handle that case here
}

try {
    // parse it
    time = format.parse(arrivalTime);
    // done
} catch (ParseException e) {
    // arrivalTime could not be parsed, you should handle that case here
}

由于日期也有routine to get the epoch-millis，您可以继续使用已有的代码，尽管在这种情况下long可能是比双重更好的返回类型。

如何在ArrayList中搜索最小数量和最短持续时间？

1 个答案: