如何將對象添加到對象哈希圖中的排序集中

Question

我正在嘗試將乘客對象添加到排序集中。 該排序的集合位於巡航對象中。 所有的巡航對象都在hashMap中。 我是一個新手，所以遇到了麻煩。 這是我嘗試做的事情。

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);

乘客的參數是字符串，字符串是姓氏，然后是他們的第一個名字首字母。 巡游的參數是日期字符串，sortedSet乘客，等待隊列的隊列和布爾值變量，以確定船只是否已出發。 運行此代碼時，我不斷收到大量錯誤。 先謝謝您的幫助。

這是我收到的錯誤。

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)

旅客艙位

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;
    }
}

郵輪班

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

發生這種情況的原因是您的passengerSet是TreeSet （ SortedSet ），這意味着它將在每次添加后對其進行排序，因為TreeSet是有序集合，並且具有不同於常規HashMap特定順序。 每個SortedMap必須知道如何對它包含的元素進行排序。 這可以通過兩種方式完成：

1. 您可以從Comparable<T>接口實現您的類。
2. 您可以將自定義Comparator<T>添加到SortedMap 。

因此，您可以通過三種方式進行修復（可能更多，但其中三種很明顯）：

1. 擺脫SortedMap ，假設在代碼中將SortedMap替換為Map並將TreeMap替換為HashMap 。

2. 將自定義比較器添加到您的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. 在Passenger類中實現Comparable<T>接口。

    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); } }