java - 如何将对象添加到对象 HashMap 中的已排序集合

Question

我正在尝试将乘客对象添加到排序集中。该排序集位于巡航对象中。所有巡航对象都在 hashMap 中。我对 Collection 有点陌生，所以遇到了麻烦。这是我尝试做的事情。

HashMap<String, Cruise> cruiseMap = new HashMap<String, Cruise>();
SortedSet<Passenger> passengerSet = new TreeSet<Passenger>();
Queue<Passenger> waitingList = new LinkedList<Passenger>();

Cruise cruise = new Cruise("1", passengerSet, waitingList, false);

cruiseMap.put("1", cruise);
Passenger passenger = new Passenger("Smith", "J");
cruiseMap.get("1").getPassengerSet().add(passenger);

乘客的参数是由姓氏和名字首字母组成的字符串。游轮的参数是字符串形式的日期、已排序的乘客、等待列表的队列和用于确定船舶是否已出发的 boolean 变量。当我运行这段代码时，我不断收到大量错误。预先感谢您的帮助。

这是我收到的错误。

Exception in thread "main" java.lang.ClassCastException: edu.ilstu.Passenger cannot be cast to java.lang.Comparable
at java.util.TreeMap.compare(Unknown Source)
at java.util.TreeMap.put(Unknown Source)
at java.util.TreeSet.add(Unknown Source)
at edu.ilstu.Driver.main(Driver.java:48)

乘客舱位

public class Passenger {
    private String lastName = "";
    private String firstName = "";

    public Passenger()
    {
        lastName = "no last name yet";
        firstName = "no first name yet";
    }
    public Passenger(String lastName, String firstName)
    {
        this.lastName = lastName;
        this.firstName = firstName;
    }
    /**
     * @return the lastName
     */
    public String getLastName()
    {
        return lastName;
    }
    /**
     * @param lastName the lastName to set
     */
    public void setLastName(String lastName)
    {
    this.lastName = lastName;
    }
    /**
     * @return the firstName
     */
    public String getFirstName()
    {
        return firstName;
    }
    /**
     * @param firstName the firstName to set
     */
    public void setFirstName(String firstName)
    {
        this.firstName = firstName;
    }
    /* (non-Javadoc)
     * @see java.lang.Object#toString()
     */
    @Override
    public String toString()
    {
        return lastName + " " + firstName;
    }
}

邮轮舱位

public class Cruise
    {
    private String day = "";
    private  SortedSet<Passenger> passengerSet = new TreeSet<Passenger>();
    private Queue<Passenger> waitingList = new LinkedList<Passenger>();
    private boolean hasDeparted = false;

    public Cruise()
    {
        day = "no day yet";
        passengerSet = null;
        waitingList = null;
        hasDeparted = false;
    }

    public Cruise(String day, SortedSet<Passenger> passengerSet,     Queue<Passenger> waitingList, boolean hasDeparted)
    {
        this.day = day;
        this.passengerSet = passengerSet;
        this.waitingList = waitingList;
        this.hasDeparted = hasDeparted;
    }

    /**
     * @return the day
     */
    public String getDay()
    {
        return day;
    }

    /**
     * @param day the day to set
     */
    public void setDay(String day)
    {
        this.day = day;
    }

    /**
     * @return the passengerSet
     */
    public SortedSet<Passenger> getPassengerSet()
    {
        return passengerSet;
    }

    /**
     * @param passengerSet the passengerSet to set
     */
    public void setPassengerSet(SortedSet<Passenger> passengerSet)
    {
        this.passengerSet = passengerSet;
    }

    /**
     * @return the waitingList
     */
    public Queue<Passenger> getWaitingList()
    {
        return waitingList;
    }

    /**
     * @param waitingList the waitingList to set
     */
    public void setWaitingList(Queue<Passenger> waitingList)
    {
        this.waitingList = waitingList;
    }

    /**
     * @return the hasDeparted
     */
    public boolean isHasDeparted()
    {
        return hasDeparted;
    }

    /**
     * @param hasDeparted the hasDeparted to set
     */
    public void setHasDeparted(boolean hasDeparted)
    {
        this.hasDeparted = hasDeparted;
    }

}

Answer 1

发生这种情况是因为您的passengerSet是 TreeSet ( SortedSet )，这意味着每次添加后它都会自行排序，因为 TreeSet是有序集并且具有与通常不同的特定顺序 HashMap 。每SortedMap必须知道如何对其包含的元素进行排序。这可以通过两种方式完成:

1. 您可以从 Comparable<T> 实现您的类(class)接口(interface)。
2. 您可以添加自定义Comparator<T>给您SortedMap .

因此，您可以通过三种方法来修复它(可能更多，但其中三种 - 是显而易见的):

1. 摆脱SortedMap ，假设替换您的 SortedMap至Map并替换TreeMap至HashMap在您的代码中。

2. 将自定义比较器添加到您的 passengerSet

   HashMap<String, Cruise> cruiseMap = new HashMap<String, Cruise>();
   SortedSet<Passenger> passengerSet = new TreeSet<Passenger>(new Comparator<Passenger>() {
   
       @Override
       public int compare(Passenger lhs, Passenger rhs) {
           return lhs.getFirstName().compareTo(rhs.getFirstName());
       }
   });
   Queue<Passenger> waitingList = new LinkedList<>();
   
   Cruise cruise = new Cruise("1", passengerSet, waitingList, false);
   
   cruiseMap.put("1", cruise);
   Passenger passenger = new Passenger("Smith", "J");
   cruiseMap.get("1").getPassengerSet().add(passenger);
   

3. 实现 Comparable<T>您的Passenger中的界面类。

   public class Passenger implements Comparable<Passenger> {
       private String lastName = "";
       private String firstName = "";
   
       public Passenger()
       {
           lastName = "no last name yet";
           firstName = "no first name yet";
       }
       public Passenger(String lastName, String firstName)
       {
           this.lastName = lastName;
           this.firstName = firstName;
       }
       /**
        * @return the lastName
        */
       public String getLastName()
       {
           return lastName;
       }
       /**
        * @param lastName the lastName to set
        */
       public void setLastName(String lastName)
       {
       this.lastName = lastName;
       }
       /**
        * @return the firstName
        */
       public String getFirstName()
       {
           return firstName;
       }
       /**
        * @param firstName the firstName to set
        */
       public void setFirstName(String firstName)
       {
           this.firstName = firstName;
       }
       /* (non-Javadoc)
        * @see java.lang.Object#toString()
        */
       @Override
       public String toString()
       {
           return lastName + " " + firstName;
       }
   
       @Override
       public int compareTo(Passenger another) {
           return firstName.compareTo(another.firstName);
       }
   }