Question

我有一个计划在多个日期交付的汽车清单，需要根据以下几点进行排序：

如果isReady>0,，则应首先在表中显示它。和那么该特定日期的其他值就会低于该值。
如果isReady>0和对象gear!=null，则它首先显示在该特定日期的表格。其次是对象gear==null所在的其他值。
如果isReady>0，对象gear!=null和对象tyre!=null，则表示该特定日期的值首先显示在表格中。其次是对象gear==null和tyre==null的其他值。

以下是class代码：

public class Car {
    private int isReady;
    private Tyre tyre;
    private Gear gear;
    private Date deliveryDate;
}


public class Gear {
    private int id;
    private String type;
}


public class Tyre {
    private int id;
    private String grip;
}

public class CarComparator implements Comparator<Car> {
    @Override
    public int compare(Car entry1, Car entry2) {
        int value = 0;

        if (entry1.getIsReady() > entry2.getIsReady()) {
            value = -1;
        } else if (entry1.getIsReady() < entry2.getIsReady()) {
            value = 1;
        } else if (entry1.getIsReady() == entry2.getIsReady()) {
            value = 0;
        }
        return value;
    }
}

我开发了一种比较器，可以在第一个条件下正常工作 isReady>0。你能帮我解决其他情况吗上面提到过。

谢谢。

Answer 1

检查此比较器，以便可以对多个属性进行排序

public class CarComparator implements Comparator<Car> {

    @Override
    public int compare(Car entry1, Car entry2) {
        int value;
        if (entry1.getDeliveryDate().before(entry2.getDeliveryDate())){
            value = -1;
        }else if (entry1.getDeliveryDate().equals(entry2.getDeliveryDate())){
            value = 0;
        }else{
            value =1;
        }
        //For same day
        if (value==0){
            if (entry1.getIsReady() > entry2.getIsReady()) {
                value = -1;
            } else if (entry1.getIsReady() < entry2.getIsReady()) {
                value = 1;
            } else if (entry1.getIsReady() == entry2.getIsReady()) {
                value = 0;
            }
        }
        //if same isReady
        if (value==0){
            if (entry1.getGear()!=null && entry2.getGear()==null) {
                value = -1;
            } else  if (entry1.getGear()==null && entry2.getGear()==null) {
                value = 0;
            } else{
                value = 1;
            }
        }
        //if still equals
        if (value==0){
            if (entry1.getTyre()!=null && entry2.getTyre()==null) {
                value = -1;
            } else  if (entry1.getTyre()==null && entry2.getTyre()==null) {
                value = 0;
            } else{
                value = 1;
            }
        }


        return value;
    }
}

我不确定这是否是您要尝试的方法。以上比较器的作用是：首先使用日期排序，如果找到相等的日期（值= 0），则比较isReady，然后比较getGear（）和最后的getTyre（）。

这样，您可以在比较器中添加所需数量的属性。

包括3辆车的主要方法

public class Main {
    public static void main (String[]args) throws UnsupportedEncodingException, ParseException {

        List<Car> carL = new ArrayList<Car>();

        Car car1 = new Car();
        car1.setDeliveryDate(new Date());
        Gear gear1 = new Gear();
        car1.setGear(gear1);
        Tyre tyre1 = new Tyre();
        car1.setTyre(null);
        car1.setId(1);
        car1.setDeliveryDate((new SimpleDateFormat("dd-MM-yyyy")).parse("01-01-2000"));
        car1.setIsReady(0);

        Car car2 = new Car();
        car2.setDeliveryDate(new Date());
        Gear gear2 = new Gear();
        car2.setGear(gear2);
        Tyre tyre2 = new Tyre();
        car2.setTyre(tyre2);
        car2.setId(2);
        car2.setDeliveryDate((new SimpleDateFormat("dd-MM-yyyy")).parse("02-01-2000"));

        car2.setIsReady(1);

        Car car3 = new Car();
        car3.setDeliveryDate(new Date());
        Gear gear3 = new Gear();
        car3.setGear(gear3);
        Tyre tyre3 = new Tyre();
        car3.setTyre(tyre3);
        car3.setId(3);
        car3.setDeliveryDate((new SimpleDateFormat("dd-MM-yyyy")).parse("01-01-2000"));

        car3.setIsReady(1);

        carL.add(car1);
        carL.add(car2);
        carL.add(car3);
        Collections.sort(carL, new CarComparator());
        for (Car car : carL) {
            System.out.println("car: " + car.toString());
        }
    }
}

输出：

car: Car{id=3, isReady=1, tyre=false, gear=false, deliveryDate=Sat Jan 01 00:00:00 EET 2000}
car: Car{id=1, isReady=0, tyre=true, gear=false, deliveryDate=Sat Jan 01 00:00:00 EET 2000}
car: Car{id=2, isReady=1, tyre=false, gear=false, deliveryDate=Sun Jan 02 00:00:00 EET 2000}

Answer 2

好吧，从Java 8开始，您可以像这样构建比较器：

//order by delivery date first, ascending order
Comparator<Car> carComparator = Comparator.comparing( Car::getDeliveryDate )
  //order by isReady in ascending order
  .thenComparing( Car::getIsReady )
  //we map null to 1 and non-null to -1 and ignore the rest for now
  .thenComparing( car -> car.getGear() != null ? -1 : 1 ) 
  .thenComparing( car -> car.getTyre() != null ? -1 : 1 );

Answer 3

说实话，您的代码没有发现任何问题。也就是说，如果您的意图是在entry1大于entry 2（规范的相反顺序）时返回-1。没有更多的代码可供阅读，我认为如果您尝试比较为您的用例构建的更大或更小的值，您的代码将可以正常工作。

但是，我认为您的退货方法效率低下。您不需要返回值。您可以返回一个实际值。

为进行比较，您可以先比较==，然后再评估其余部分。但这可能很难阅读代码，所以我给你两个版本。

删除值版本：

public class CarComparator implements Comparator<Car> {
    @Override
    public int compare(Car entry1, Car entry2) {
        if (entry1.getisReady() > entry2.getisReady()) {
            return -1;
        } else if (entry1.getisReady() < entry2.getisReady()) {
            return 1;
        } else if (entry1.getisReady() == entry2.getisReady()) {
            return 0;
        }
    }
}

去除价值和不同的比较方式：

public class CarComparator implements Comparator<Car> {

    @Override
    public int compare(Car entry1, Car entry2) {
        if (entry1.getisReady() == entry2.getisReady()) return 0;            
        return entry1.getisReady() > entry2.getisReady()? -1 : 1;
    }
}

我不确定我是否正确理解了您，但是如果我没有正确理解，希望此代码可以为您提供帮助。我将它们分为3种方法，希望您得到一个getTyre（）和getGear（）方法。您可以根据需要将它们组合起来，对于最后一种方法，其值被分成可变的顺序以便于阅读代码。

class CarComparator implements Comparator<Car> {

    public int compare(Car entry1, Car entry2) {
        if (entry1.getisReady() == entry2.getisReady()) return 0;
        return entry1.getisReady() > entry2.getisReady()? -1 : 1;
    }

    public int compareGear(Car entry1, Car entry2){
        if ( (entry1.getGear() != null && entry2.getGear() != null) 
           ||(entry1.getGear() == null && entry2.getGear() == null)
           ){
            return compare(entry1, entry2);
        }  
        return entry1.getGear() != null && entry2.getGear() == null? -1 : 1;

    }

    public int compareTye(Car entry1, Car entry2){
        int order1 = entry1.getGear() != null && entry1.getTyre() != null? 1 : 0;
        int order2 = entry2.getGear() != null && entry2.getTyre() != null? 1 : 0;

        if ( order1 == order2 ) return compare(entry1, entry2);
        return order1 > order2? -1 : 1;
    } 
}

Answer 4

为什么不重用Integer.compareTo来使代码更短？

像这样：

import java.util.Comparator;

public class CarComparator implements Comparator<Car> {
    @Override
    public int compare(Car entry1, Car entry2) {
        int value = 0;

        // might want to add a null check for either entry1 and entry2

        value = entry1.getDeliveryDate().compareTo(entry2.getDeliveryDate());
        if (value == 0) {
            value = ((Integer)entry1.getIsReady()).compareTo((Integer)entry2.getIsReady());
            if (value == 0) {
                value = getIntegerValueForNullCheck(entry1.getGear()).compareTo(getIntegerValueForNullCheck(entry2.getGear()));
                if (value == 0) {
                    value = getIntegerValueForNullCheck(entry1.getTyre()).compareTo(getIntegerValueForNullCheck(entry2.getTyre()));
                }
            }
        }

        return value;
    }
    private Integer getIntegerValueForNullCheck (Object o) {
        return o == null ? 0 : 1;
    }
}

包括测试排序的代码：

import java.time.Duration;
import java.time.Instant;
import java.time.LocalDateTime;
import java.util.*;

public class Sorting {

    public static void main(String[] args) {

        List<Car> cars = new LinkedList<>();


        Date today = new Date();
        Instant now = Instant.now();
        Instant after = now.plus(Duration.ofDays(1));
        Date tomorrow = Date.from(after);

        cars.add(new Car(5, new Tyre(1,"1"), new Gear(1, "1"), today ));
        cars.add(new Car(5, new Tyre(1,"1"), null, today ));
        cars.add(new Car(5, null, null, today ));
        cars.add(new Car(4, null, null, today ));
        cars.add(new Car(3, null, null, tomorrow ));


        Collections.sort(cars, new CarComparator());
        System.out.println(cars);
    }

}

输出：

[Car{isReady=4, tyre=null, gear=null, deliveryDate=Thu Oct 10 11:27:20 IDT 2019}
, Car{isReady=5, tyre=null, gear=null, deliveryDate=Thu Oct 10 11:27:20 IDT 2019}
, Car{isReady=5, tyre=Tyre{id=1, grip='1'}, gear=null, deliveryDate=Thu Oct 10 11:27:20 IDT 2019}
, Car{isReady=5, tyre=Tyre{id=1, grip='1'}, gear=Gear{id=1, type='1'}, deliveryDate=Thu Oct 10 11:27:20 IDT 2019}
, Car{isReady=3, tyre=null, gear=null, deliveryDate=Fri Oct 11 11:27:20 IDT 2019}
]

根据Java中的嵌套值对对象列表进行排序

4 个答案: