求java 单链表基本 *** 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

//卜禅测试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退型梁尘出)：")

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("删除当前渣毕链表中 " + 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=headi>0 &&pre.getNext()!=nulli--,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=headi>0 &&pre.getNext()!=nulli--,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!=nullpos++,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!=nullcurr=curr.getNext()){

System.out.print(curr.getData() + "\t")

}

System.out.println()

}

}

java.util.Linkedlist是双弯袜向链表，当然也就包配闹埋括了单链表的功能，你可以去看他怎么写的啊

public class SingleLinkedList<E>{

private Entry<E>first, last

private int size = 0

public void add(E element) {

Entry<E>newEntry = new Entry<培蚂E>(element, null)

if (first == null) {

first = last = newEntry

} else {

last.next = newEntry

last = newEntry

}

++size

}

public E get(int index) {

if (index <0 || index >= size)

throw new IndexOutOfBoundsException("Index: "+index+

", Size: "+size)

Entry<E>e = first

for (int i = 0i <index++i)

e = e.next

return e.data

}

private static class Entry<E>{

Entry(E data, Entry<E>next) {

this.data = data

this.next = next

}

E data

Entry<E>next

}

}

public class Link {

Node head = null

Node point = null

Node newNode = null

public int Count = 0//统计值

//插入

public void AddNode(int t) {

newNode = new Node()

if (head == null) {

head = newNode

} else {

point = head

while (point.next != null) {

point = point.next

}

point.next = newNode

}

point = newNode

point.vlaue = t

point.next = null

Count++

}

//返回裂谨值

public int GetValue(int i) {

if (head == null || i <0 || i >Count)

return -999999

int n

Node temp = null

point = head

for (n = 0n <= in++) {

temp = point

point = point.next

}

return temp.vlaue

}

//删除

public void DeleteNode(int i) {

if (i <肆陪基 0 || i >Count) {

return

}

if (i == 0) {

head = head.next

} else {

int n = 0

point = head

Node temp = point

for (n = 0n <in++) {

temp = point

point = point.next

}

temp.next = point.next

}

Count--

}

//排序

public void Sotr() {

for (Node i = headi != nulli = i.next) {

for (Node j = i.nextj != nullj = j.next) {

if (i.vlaue >乱陵 j.vlaue) {

int t = i.vlaue

i.vlaue = j.vlaue

j.vlaue = t

}

}

}

}

}

class Node {

int vlaue

Node next

}

---