使用compareTo（）方法按字母顺序对列表进行排序

Question

我正在用java编写电话簿程序，我需要按字母顺序列出列表中的人员，为此我需要为java中的列表编写排序算法，它应该只使用compareTo（）方法。 所以有人可以帮助我做到这一点吗？

public void listAlpha()
 {
     Node tempNode = head;

     for(int i = 0; i <= size; i++)
        {
            for(int j = 0; j <= i; j++)
            {
                int comparison = ((tempNode.getNext().getElement().getName()).compareTo(tempNode.getElement().getName()));
                if(comparison < 0)
                {
                    Person tempPerson =  tempNode.getElement();

                    tempNode.setElement(tempNode.getNext().getElement());
                    tempNode.getNext().setElement(tempPerson);
                    tempNode = tempNode.getNext();
                }
            }


        }

（顺便说一下，这是一个家庭作业，我使用自己的数据结构。）

这是我上面写的方法所属的类：

import java.util.*;

/** Singly linked list implementation .*/
public class SLinkedList<E> implements LinkedList<E>, Iterable<E> {
  protected Node<E> head;       // head node of the list
  protected Node<E> tail;       // tail node of the list
  protected int size;       // number of nodes in the list

    public Iterator<E> iterator()
    {
        return new LinkedListIterator(head);
    }

  /** Default constructor that creates an empty list */
  public SLinkedList() {
    head = null;
    tail = null;
    size = 0;
  }

  public int size() { 
    return size;
  }

  public boolean isEmpty() {
    return size == 0;
  }

  public void addFirst(E newElement) {
    Node<E> newNode = new Node(newElement,null);
    if(size == 0) //if list is empty
        tail = newNode;

    newNode.setNext(head);
    head = newNode;
    size++;
  }

  public void addLast(E newElement) {
    Node<E> newNode = new Node(newElement,null);

    if(size == 0) //if list is empty
        head = newNode;

    if (size != 0) //if list is not empty
        tail.setNext(newNode);

    tail = newNode;
    size++;
  }

  public E removeFirst() {
    Node<E> tempNode = null;
    if (size != 0) {
        if(size == 1)
            tail = null;

        tempNode = head;
        head = head.getNext();
        tempNode.setNext(null);
        size--;
    }

    //if list is empty then return null
    return tempNode.getElement(); 

  }

  public E removeLast() {
    Node<E> tempNode = head;

    if(size == 0)
        return null;

    if(size == 1) {
        head = null;
        tail = null;
        size--;
        return tempNode.getElement();
    }

    //size is greater than 1
    for(int i=1; i<=size-2; i++) {
        tempNode = tempNode.getNext(); //go to element that before the tail
    }

    Node<E> tempNode2 = tail;
    tail = tempNode;
    tail.setNext(null);
    size--;
    return tempNode2.getElement();

  }

  public void remove(E element){
    int index = 0;
    boolean found = false;
    Node<E> temp = head;
    for(int i=1; i<=size; i++) {//find the node with element
        index++;
        if(temp.getElement().equals(element)){
            found = true;
            break;
        }
        temp = temp.getNext(); 
    }

    if(found){
        if(index == 1) 
            removeFirst();

        else if(index == size)
            removeLast();

        else{   
            //find the previous node
            Node<E> prev = head;
            for(int i=1; i<index-1; i++) {
                prev = prev.getNext(); 
            }

            prev.setNext(temp.getNext());
            temp.setNext(null);
            size--;
        }
    }
  }

  public int searchList(E searchKey) {
    if(size == 0)
        return -1;

    Node tempNode = head;
    for(int i=1; i<=size; i++) {
        if(tempNode.getElement().equals(searchKey))
            return i; //return index of the node
        tempNode = tempNode.getNext();
    }

    return -1; //not found
  }

  public void printList() {
    Node tempNode = head;
    for(int i=1; i<=size; i++) {
        System.out.print(tempNode.getElement());
        if(i!=size) //if it is not last element
            System.out.print(" - ");
        tempNode = tempNode.getNext();
    }
    System.out.println();

  }

人员类：

public class Person
{
    private String name;
    private String surname;
    private String address;
    private PhoneNumber phone1;
    private PhoneNumber phone2;
    private PhoneNumber phone3;

    public Person()
    {
        name = null;
        surname = null;
        address = null;
        phone1.setPhone(0);
        phone1.setType("");
        phone2.setPhone(0);
        phone2.setType("");
        phone3.setPhone(0);
        phone3.setType("");
    }

    public Person(String n, String s, String a,PhoneNumber p1, PhoneNumber p2, PhoneNumber p3)
    {
        name = n;
        surname = s;
        address = a;
        phone1 = p1;
        phone2 = p2;
        phone3 = p3;

    }

    public String getName()
    {
        return name;
    }

    public void setName(String n)
    {
        name = n;
    }
    public String getSur()
    {
        return surname;
    }

    public void setSur(String s)
    {
        surname = s;
    }

    public void insertPhone(PhoneNumber phone)
    {
        if(phone2 == null)
            phone2 = phone;
        else if(phone3 == null)
            phone3 = phone;
    }

    public PhoneNumber getPhone1()
    {
        return phone1;
    }

    public PhoneNumber getPhone2()
    {
        return phone2;
    }

    public PhoneNumber getPhone3()
    {
        return phone3;
    }

    public String getAdd()
    {
        return address;
    }

    public void setAdd(String a)
    {
        address = a;
    }

Answer 1

您可以使Person类实现Comparable ，并定义以下方法：

    public class Person implements Comparable<Person> {

       // Your previous code

       public int compareTo(Person other) {
          if (other == null) {
             // throw exception for example
          }
          return this.name.toLowerCase().compareTo(other.name.toLowerCase());
       }
    }

Answer 2

正如其他人所提到的， compareTo是Comparable接口的一部分。

如何实现它取决于您是想先按姓氏或名字排序，还是希望按升序排序。

例如，如果您想先按姓氏排序，则按升序排序：

public class Person implements Comparable<Person> {
    // the parts of Person you already have would go here
    public int compareTo(Person person) {
        if (person == null) {
            return -1;
        }

        if (surname != null && person.getSur() == null) {
            return -1;
        } else if (surname == null && person.getSur() != null) {
            return 1;
        } else if (surname != null && person.getSur() != null) {
            int compare = surname.compareToIgnoreCase(person.getSur());
            if (compare != 0) {
                return compare;
            }
        }
        // Note that we did nothing if both surnames were null or equal

        if (name == null && person.getName() == null) {
            return 0;
        } else if (name != null && person.getName() == null) {
            return -1;
        } else if (name == null && person.getName() != null) {
            return 1;
        } else {
            return name.compareToIgnoreCase(person.getName());
        }
    }
}

（我实际上没有测试这段代码）

这依赖于String的compareToIgnoreCase实现。

请注意，这也会将具有空名称和姓氏的所有空对象和对象移动到列表的末尾。

说了这么多，如果你实现了Comparable，你可以使用sort来使Collections API为你工作。

如果发现对象需要多种不同的排序方法，则可以创建一组Comparator对象来进行排序。

Answer 3

在Person类中实现Comparable 。

您的compareTo（）方法将类似于：

public int compareTo(Person other) {
    return name.compareTo(other.getName())
}

然后使用Collections.sort(<your list of Person>);

Answer 4

Person类的签名应该是这样的：

public class Person implements Comparable<Person>

将compareTo-method添加到Person类，并使用Collections.sort（personList）作为starf建议。