求java 单链表基本操作的实现

/*
注意：链表的结点数量用size表示，结点的位置为0~size-1
*/
import java.util.Scanner;

public class Test7 {
public static void main(String[] args) {
try{
LinkList list = new LinkList();
Integer value;
int pos = 0;
Scanner input = new Scanner(System.in);
String choice = null;

//测试A
while(true){
System.out.print("请输入待插入结点的值(x或X退出)：");
choice = input.next();
if(choice.toUpperCase().equals("X")){
break;
}
value = Integer.valueOf(choice);
if(list.addAt(pos, value) == true){
System.out.println("插入值为 " + value + " 的结点到当前链表成功!");
pos++;
}
else{
System.out.println("插入结点失败!");
}
}

System.out.print("当前链表所有结点：");
list.listAll();

//测试B
while(true){
System.out.print("请输入待查询结点的值(x或X退出)：");
choice = input.next();
if(choice.toUpperCase().equals("X")){
break;
}
value = Integer.valueOf(choice);
pos = list.findByValue(value);
if(pos == -1){
System.out.println("当前链表中不存在值为 " + value + " 的结点");
}
else{
System.out.println("值为 " + value + " 的结点在当前链表中的位置为 " + pos);
}
}

//测试C
while(true){
System.out.print("请输入待删除结点的位置[0~" + (list.getSize()-1) + "](x或X退出)：");
choice = input.next();
if(choice.toUpperCase().equals("X")){
break;
}
pos = Integer.valueOf(choice);
if(list.removeAt(pos) == true){
System.out.println("删除当前链表中 " + pos + " 位置的结点成功!");
}
else{
System.out.println("删除结点失败!");
}
}

System.out.print("当前链表所有结点：");
list.listAll();
}
catch(Exception e){
e.printStackTrace();
}
}
}

/**
* 链表结点类
*/
class Node{
private Object data; //链表结点的数据域
private Node next; //链表结点的指针域，指向直接后继结点

public Node(){
data = null;
next = null;
}

public Node(Object data, Node next){
this.data = data;
this.next = next;
}

public Object getData(){
return this.data;
}

public void setData(Object data){
this.data = data;
}

public Node getNext(){
return this.next;
}

public void setNext(Node next){
this.next = next;
}
}

/**
* 链表类
*/
class LinkList{
private Node head = null; //头结点指针
private int size = 0;

public LinkList(){
head = new Node();
size = 0;
}

//在i位置插入元素elem
public boolean addAt(int i, Object elem) {
if(i < 0 || i > size){
return false;
}

Node pre,curr;
int pos;
for(pre=head; i>0 && pre.getNext()!=null; i--,pre=pre.getNext());
curr = new Node(elem, pre.getNext());
pre.setNext(curr);
size++;
return true;
}

//删除i位置的元素
public boolean removeAt(int i) {
if(i < 0 || i >= size){
return false;
}

Node pre,curr;
for(pre=head; i>0 && pre.getNext()!=null; i--,pre=pre.getNext());
curr = pre.getNext();
pre.setNext(curr.getNext());
size--;
return true;
}

//根据值value查询结点是否存在，若存在返回位置，否则返回-1
public int findByValue(Object value){
Node curr;
int pos;
for(pos=0,curr=head.getNext(); curr!=null; pos++,curr=curr.getNext()){
if(curr.getData().toString().equals(value.toString())){
break;
}
}

if(curr==null){
return -1;
}
return pos;
//return (curr!=null ? pos : -1);
}

public int getSize(){
return size;
}

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

public void listAll(){
for(Node curr=head.getNext(); curr!=null; curr=curr.getNext()){
System.out.print(curr.getData() + "\t");
}
System.out.println();
}
}