我一直在问关于创建一个排序方法来排序从一个文本文件,并已从我以前的问题:(What is a better method to sort strings alphabetically in a linked list that is reading in lines from a text file?),先进的,但现在运行到两个错误后创建一个名为AlphaSorter.java与方法compareTo()旨在重写我的排序()方法在我的链接。表程序/类调用ContactList.java如何解决此错误:“AlphaSorter必须实现继承的抽象方法java.lang.Comparable <AlphaSorter> .compareTo(AlphaSorter)?
我收到两个错误:
AlphaSorter必须实现继承抽象方法java.lang.Comparable.compareTo(AlphaSorter)
类型AlphaSorter的方法的compareTo(ContactNode,ContactNode)必须重写或实现的超类型方法
这是链表程序:ContactList.java
public class ContactList{
private ContactNode head;
private ContactNode last;
public ContactNode current;
public ContactList(){
head = null;
last = null;
current = null;}
public void addNode(ContactNode input){
if(this.head == null)
{this.head = input;
this.last = input;}
else
last.setNext(input);
input.setPrev(last);
this.last = input;}
public void traverse(){
System.out.println();
current = this.head;
while (current != null){
System.out.print(current.getName() + " ");
System.out.println("");
current = current.getNext();}
System.out.println();}
@Override
public String toString(){
ContactNode current = head;
while(current!=null){
System.out.print(current.getName() + "\n");
current = current.getNext();}
return null;}
public void insertNewFirstNode(String current){
ContactNode newNode = new ContactNode(current);
head = newNode;
if(last == null){
last = head;}}
public void sort(){
ContactList sorted = new ContactList();
ContactNode current = head
while (current != null){
if((current.getName() != null)){
current.getName().compareTo(current.getName());
sorted.insertNewFirstNode(current.getName());}
else if((current != null)){current = current.getNext();}
System.out.println(toString() + sorted);
System.out.println("");
System.out.println("");
break;}}}
这是节点类:ContactNode.java
public class ContactNode{
public String name;
public int index;
private ContactNode prev;
public ContactNode next;
ContactNode(String a){
this.name = a;
index = 0;
next = null;
prev = null;}
ContactNode(){}
public ContactNode getNext()
{return next;}
public ContactNode getPrev()
{return prev;}
public String getName()
{return name;}
public int getIndex(){
return index;}
public void setNext(ContactNode newnext)
{next = newnext;}
public void setPrev(ContactNode newprevious)
{prev = newprevious;}
public void setName(String a)
{name=a;}
public void setIndex(int b)
{index=b;}}
这是主要的方法:ContactMain.java
import java.util.Scanner;
import java.io.FileReader;
import java.io.FileNotFoundException;
public class ContactMain{
public static void main(String[]args){
try{
FileReader filepath = new FileReader("data1.txt");
Scanner k = new Scanner(filepath);
ContactList myList = new ContactList();
while (k.hasNextLine()){
String i = k.nextLine();
myList.addNode(new ContactNode(i));}
myList.traverse();
System.out.println("");
myList.sort();
}catch (FileNotFoundException e){
System.out.println("File Not Found. ");}}}
最后是这样的排序类:AlphaSorter.java
import java.util.Comparator;
import java.io.Serializable;
class AlphaSorter implements Comparable<AlphaSorter>{
@Override
public int compareTo(ContactNode e1, ContactNode e2) {
return e1.getName().compareTo(e2.getName());}
public boolean equals(Object obj){
return this==obj;}}
我不确定的的CompareTo()方法是如何准确地与工作导入包。有人可以解释吗?谢谢!
很好。这解决了我的错误,但我的输出仍然是我的原始列表的副本。如何按字母顺序执行? –