我们获得了从头开始创建LinkedList的任务,并且绝对没有给出读数来指导我们这个由migrane引起的任务。此外,所有在线似乎只使用Java内置的LinkedList方法和东西。无论如何,链接列表在使用Java的默认内容时非常有意义,但是从头开始创建它没有任何意义。可以说我有
public class LinkedList {
private LinkedList next;
private final String word;
// constructor
public LinkedList(String word, LinkedList next) {
this.word = word;
this.next = next;
}
因此神奇地我们有一个链表。到底是怎么回事?我是如何创建这样的链接列表的?这是如何运作的?我应该编写一个append方法,将给定的String word
参数添加到this
链表的末尾。我试着看看内置java链接列表类的addLast内置方法,但它对我没有帮助,因为我真的不明白发生了什么。任何人都想帮助我:)
答案 0 :(得分:42)
如果您实际上正在构建一个真实的系统,那么是的,如果您需要的话,那么您通常只需使用标准库中的内容。也就是说,不要认为这是毫无意义的运动。理解事物是如何工作的很好,理解链表是理解更复杂数据结构的重要一步,其中许多数据结构在标准库中不存在。
您创建链接列表的方式与Java集合API执行方式之间存在一些差异。 Collections API试图遵循更复杂的界面。 Collections API链表也是一个双向链表,而您正在构建一个单链表。你正在做什么更适合课堂作业。
使用LinkedList
类,实例将始终是至少一个元素的列表。通过这种设置,您可以在需要空列表时使用null
。
将next
视为“列表的其余部分”。实际上,许多类似的实现使用名称“tail”而不是“next”。
以下是包含3个元素的LinkedList
图表:
请注意,它是指向单词(“Hello”)的LinkedList
对象和2个元素的列表。 2个元素的列表有一个单词(“Stack”)和一个1元素的列表。 1个元素的列表有一个单词(“溢出”)和一个空列表(null
)。因此,您可以将next
视为另一个恰好是一个元素更短的列表。
您可能想要添加另一个只接受String的构造函数,并在null
旁边设置。这将用于创建1元素列表。
要追加,请检查next
是否为null
。如果是,请创建一个新的元素列表并将next
设置为该列表。
next = new LinkedList(word);
如果下一个不是null
,则改为追加next
。
next.append(word);
这是递归方法,这是最少量的代码。您可以将其转换为在Java * 中更有效的迭代解决方案,并且不会冒很长列表的堆栈溢出的风险,但我猜测不需要复杂程度为你的任务。
*有些语言有尾调用消除,这是一种优化,它允许语言实现将“尾调用”(调用另一个函数作为返回前的最后一步)转换为(有效)“goto” 。这使得这样的代码完全避免使用堆栈,这使得它更安全(如果不使用堆栈,则不能溢出堆栈)并且通常更有效。 Scheme可能是具有此功能的语言中最着名的示例。
答案 1 :(得分:23)
您编码的内容不是LinkedList,至少不是我认可的内容。对于此分配,您需要创建两个类:
LinkNode
LinkedList
LinkNode
有一个成员字段用于其包含的数据,LinkNode
引用LinkNode
中的下一个LinkedList
。是的,它是一个自引用数据结构。 LinkedList
只有一个特殊的LinkNode
引用,引用列表中的第一个项目。
当您在LinkedList
中添加项目时,您将遍历所有LinkNode's
,直至到达最后一个项目。这个LinkNode's
接下来应该为null。然后,您在此处构建新的LinkNode
,设置其值,并将其添加到LinkedList
。
public class LinkNode {
String data;
LinkNode next;
public LinkNode(String item) {
data = item;
}
}
public class LinkedList {
LinkNode head;
public LinkedList(String item) {
head = new LinkNode(item);
}
public void add(String item) {
//pseudo code: while next isn't null, walk the list
//once you reach the end, create a new LinkNode and add the item to it. Then
//set the last LinkNode's next to this new LinkNode
}
}
答案 2 :(得分:11)
非递归链接列表的全功能实现怎么样?
我为我的Algorithms I类创建了这个作为跳板的基石,以便在为编配编写双向链接队列类之前获得更好的理解。
以下是代码:
import java.util.Iterator;
import java.util.NoSuchElementException;
public class LinkedList<T> implements Iterable<T> {
private Node first;
private Node last;
private int N;
public LinkedList() {
first = null;
last = null;
N = 0;
}
public void add(T item) {
if (item == null) { throw new NullPointerException("The first argument for addLast() is null."); }
if (!isEmpty()) {
Node prev = last;
last = new Node(item, null);
prev.next = last;
}
else {
last = new Node(item, null);
first = last;
}
N++;
}
public boolean remove(T item) {
if (isEmpty()) { throw new IllegalStateException("Cannot remove() from and empty list."); }
boolean result = false;
Node prev = first;
Node curr = first;
while (curr.next != null || curr == last) {
if (curr.data.equals(item)) {
// remove the last remaining element
if (N == 1) { first = null; last = null; }
// remove first element
else if (curr.equals(first)) { first = first.next; }
// remove last element
else if (curr.equals(last)) { last = prev; last.next = null; }
// remove element
else { prev.next = curr.next; }
N--;
result = true;
break;
}
prev = curr;
curr = prev.next;
}
return result;
}
public int size() {
return N;
}
public boolean isEmpty() {
return N == 0;
}
private class Node {
private T data;
private Node next;
public Node(T data, Node next) {
this.data = data;
this.next = next;
}
}
public Iterator<T> iterator() { return new LinkedListIterator(); }
private class LinkedListIterator implements Iterator<T> {
private Node current = first;
public T next() {
if (!hasNext()) { throw new NoSuchElementException(); }
T item = current.data;
current = current.next;
return item;
}
public boolean hasNext() { return current != null; }
public void remove() { throw new UnsupportedOperationException(); }
}
@Override public String toString() {
StringBuilder s = new StringBuilder();
for (T item : this)
s.append(item + " ");
return s.toString();
}
public static void main(String[] args) {
LinkedList<String> list = new LinkedList<>();
while(!StdIn.isEmpty()) {
String input = StdIn.readString();
if (input.equals("print")) { StdOut.println(list.toString()); continue; }
if (input.charAt(0) == ('+')) { list.add(input.substring(1)); continue; }
if (input.charAt(0) == ('-')) { list.remove(input.substring(1)); continue; }
break;
}
}
}
注意:这是单链表的非常基本的实现。 'T'类型是通用类型占位符。基本上,此链接列表应该适用于从Object继承的任何类型。如果将它用于基本类型,请确保使用可等级的等价类(对于'int'类型,请使用'Integer')。 'last'变量不是必需的,只是它将插入时间缩短为O(1)。删除速度很慢,因为它们在O(N)时间内运行,但它允许您删除列表中第一次出现的值。
如果您愿意,还可以考虑实施:
老实说,只需要几行代码就可以使它成为一个双向链表。这与双链表之间的主要区别在于双链表的节点实例需要一个指向列表中前一个元素的附加引用。
这对于递归实现的好处是它更快,并且您在遍历大型列表时不必担心泛滥堆栈。
在调试器/控制台中有3个命令可以测试它:
如果您从未见过其中一个如何工作的内部建议,我建议您在调试器中逐步执行以下操作:
虽然对于像列表这样的列表有更好更有效的方法,但了解应用程序如何通过引用/指针遍历是理解有多少高级数据结构工作的必要条件。
答案 3 :(得分:8)
提示1:阅读http://en.wikipedia.org/wiki/Linked_list
处的链接列表说明提示2:LinkedList的Java实现是一个双向链表。你的是一个单独的链表。算法不直接适用。
此外:
...但是从头开始创建[链表类]没有任何意义。
这取决于所需的工作成果。如果目标是生成满足某些功能/非功能要求的代码,那么您是对的。如果真正的目标是学习如何编程/设计API /实现非平凡的数据结构,那么最终产品的效用几乎完全无关紧要。< / p>
因此神奇地我们有一个链表
您实际拥有的是一种开放数据类型,可用于构建(某种)列表。但这不是你老师想要的。它肯定不会被认为是有用的列表抽象。一个有用的抽象包括:
执行程序员不希望一遍又一遍地重复的事情的方法,
一个抽象层,阻止程序员“破坏”列表;例如意外地创建一个循环,或意外地将子列表拼接在两个列表中以创建倒置树。
答案 4 :(得分:5)
当然,链接列表对编程n00bs有点混乱,几乎所有的诱惑都是将它看作俄罗斯玩偶,因为它看起来像是LinkedList对象中的LinkedList对象。但这很难想象,而是像计算机那样看待它。
LinkedList =数据+下一个成员
如果next是NULL
,它是列表的最后一个成员所以5个成员的LinkedList将是:
LinkedList(Data1,LinkedList(Data2,LinkedList(Data3,LinkedList(Data4,LinkedList(Data5,NULL)))))
但你可以简单地想到它:
Data1 - &gt;数据2 - &gt;数据3 - &gt; Data4 - &gt;数据5 - &gt; NULL
那么,我们如何找到这个结束?好吧,我们知道NULL就是结束所以:
public void append(LinkedList myNextNode) {
LinkedList current = this; //Make a variable to store a pointer to this LinkedList
while (current.next != NULL) { //While we're not at the last node of the LinkedList
current = current.next; //Go further down the rabbit hole.
}
current.next = myNextNode; //Now we're at the end, so simply replace the NULL with another Linked List!
return; //and we're done!
}
这当然是一个非常简单的代码,如果你给它一个循环链表,将无限循环!但那是基础。
答案 5 :(得分:4)
具有以下功能的链接列表程序
1 Insert At Start
2 Insert At End
3 Insert At any Position
4 Delete At any Position
5 Display
6 Get Size
7 Empty Status
8 Replace data at given postion
9 Search Element by position
10 Delete a Node by Given Data
11 Search Element Iteratively
12 Search Element Recursively
package com.elegant.ds.linkedlist.practice;
import java.util.Scanner;
class Node {
Node link = null;
int data = 0;
public Node() {
link = null;
data = 0;
}
public Node(int data, Node link) {
this.data = data;
this.link = null;
}
public Node getLink() {
return link;
}
public void setLink(Node link) {
this.link = link;
}
public int getData() {
return data;
}
public void setData(int data) {
this.data = data;
}
}
class SinglyLinkedListImpl {
Node start = null;
Node end = null;
int size = 0;
public SinglyLinkedListImpl() {
start = null;
end = null;
size = 0;
}
public void insertAtStart(int data) {
Node nptr = new Node(data, null);
if (start == null) {
start = nptr;
end = start;
} else {
nptr.setLink(start);
start = nptr;
}
size++;
}
public void insertAtEnd(int data) {
Node nptr = new Node(data, null);
if (start == null) {
start = nptr;
end = nptr;
} else {
end.setLink(nptr);
end = nptr;
}
size++;
}
public void insertAtPosition(int position, int data) {
Node nptr = new Node(data, null);
Node ptr = start;
position = position - 1;
for (int i = 1; i < size; i++) {
if (i == position) {
Node temp = ptr.getLink();
ptr.setLink(nptr);
nptr.setLink(temp);
break;
}
ptr = ptr.getLink();
}
size++;
}
public void repleaceDataAtPosition(int position, int data) {
if (start == null) {
System.out.println("Empty!");
return;
}
Node ptr = start;
for (int i = 1; i < size; i++) {
if (i == position) {
ptr.setData(data);
}
ptr = ptr.getLink();
}
}
public void deleteAtPosition(int position) {
if (start == null) {
System.out.println("Empty!");
return;
}
if (position == size) {
Node startPtr = start;
Node endPtr = start;
while (startPtr != null) {
endPtr = startPtr;
startPtr = startPtr.getLink();
}
end = endPtr;
end.setLink(null);
size--;
return;
}
Node ptr = start;
position = position - 1;
for (int i = 1; i < size; i++) {
if (i == position) {
Node temp = ptr.getLink();
temp = temp.getLink();
ptr.setLink(temp);
break;
}
ptr = ptr.getLink();
}
size--;
}
public void deleteNodeByGivenData(int data) {
if (start == null) {
System.out.println("Empty!");
return;
}
if (start.getData() == data && start.getLink() == null) {
start = null;
end = null;
size--;
return;
}
if (start.getData() == data && start.getLink() != null) {
start = start.getLink();
size--;
return;
}
if (end.getData() == data) {
Node startPtr = start;
Node endPtr = start;
startPtr = startPtr.getLink();
while (startPtr.getLink() != null) {
endPtr = startPtr;
startPtr = startPtr.getLink();
}
end = endPtr;
end.setLink(null);
size--;
return;
}
Node startPtr = start;
Node prevLink = startPtr;
startPtr = startPtr.getLink();
while (startPtr.getData() != data && startPtr.getLink() != null) {
prevLink = startPtr;
startPtr = startPtr.getLink();
}
if (startPtr.getData() == data) {
Node temp = prevLink.getLink();
temp = temp.getLink();
prevLink.setLink(temp);
size--;
return;
}
System.out.println(data + " not found!");
}
public void disply() {
if (start == null) {
System.out.println("Empty!");
return;
}
if (start.getLink() == null) {
System.out.println(start.getData());
return;
}
Node ptr = start;
System.out.print(ptr.getData() + "->");
ptr = start.getLink();
while (ptr.getLink() != null) {
System.out.print(ptr.getData() + "->");
ptr = ptr.getLink();
}
System.out.println(ptr.getData() + "\n");
}
public void searchElementByPosition(int position) {
if (position == 1) {
System.out.println("Element at " + position + " is : " + start.getData());
return;
}
if (position == size) {
System.out.println("Element at " + position + " is : " + end.getData());
return;
}
Node ptr = start;
for (int i = 1; i < size; i++) {
if (i == position) {
System.out.println("Element at " + position + " is : " + ptr.getData());
break;
}
ptr = ptr.getLink();
}
}
public void searchElementIteratively(int data) {
if (isEmpty()) {
System.out.println("Empty!");
return;
}
if (start.getData() == data) {
System.out.println(data + " found at " + 1 + " position");
return;
}
if (start.getLink() != null && end.getData() == data) {
System.out.println(data + " found at " + size + " position");
return;
}
Node startPtr = start;
int position = 0;
while (startPtr.getLink() != null) {
++position;
if (startPtr.getData() == data) {
break;
}
startPtr = startPtr.getLink();
}
if (startPtr.getData() == data) {
System.out.println(data + " found at " + position);
return;
}
System.out.println(data + " No found!");
}
public void searchElementRecursively(Node start, int data, int count) {
if (isEmpty()) {
System.out.println("Empty!");
return;
}
if (start.getData() == data) {
System.out.println(data + " found at " + (++count));
return;
}
if (start.getLink() == null) {
System.out.println(data + " not found!");
return;
}
searchElementRecursively(start.getLink(), data, ++count);
}
public int getSize() {
return size;
}
public boolean isEmpty() {
return start == null;
}
}
public class SinglyLinkedList {
@SuppressWarnings("resource")
public static void main(String[] args) {
SinglyLinkedListImpl listImpl = new SinglyLinkedListImpl();
System.out.println("Singly Linked list : ");
boolean yes = true;
do {
System.out.println("1 Insert At Start :");
System.out.println("2 Insert At End :");
System.out.println("3 Insert At any Position :");
System.out.println("4 Delete At any Position :");
System.out.println("5 Display :");
System.out.println("6 Get Size");
System.out.println("7 Empty Status");
System.out.println("8 Replace data at given postion");
System.out.println("9 Search Element by position ");
System.out.println("10 Delete a Node by Given Data");
System.out.println("11 Search Element Iteratively");
System.out.println("12 Search Element Recursively");
System.out.println("13 Exit :");
Scanner scanner = new Scanner(System.in);
int choice = scanner.nextInt();
switch (choice) {
case 1:
listImpl.insertAtStart(scanner.nextInt());
break;
case 2:
listImpl.insertAtEnd(scanner.nextInt());
break;
case 3:
int position = scanner.nextInt();
if (position <= 1 || position > listImpl.getSize()) {
System.out.println("invalid position!");
} else {
listImpl.insertAtPosition(position, scanner.nextInt());
}
break;
case 4:
int deletePosition = scanner.nextInt();
if (deletePosition <= 1 || deletePosition > listImpl.getSize()) {
System.out.println("invalid position!");
} else {
listImpl.deleteAtPosition(deletePosition);
}
break;
case 5:
listImpl.disply();
break;
case 6:
System.out.println(listImpl.getSize());
break;
case 7:
System.out.println(listImpl.isEmpty());
break;
case 8:
int replacePosition = scanner.nextInt();
if (replacePosition < 1 || replacePosition > listImpl.getSize()) {
System.out.println("Invalid position!");
} else {
listImpl.repleaceDataAtPosition(replacePosition, scanner.nextInt());
}
break;
case 9:
int searchPosition = scanner.nextInt();
if (searchPosition < 1 || searchPosition > listImpl.getSize()) {
System.out.println("Invalid position!");
} else {
listImpl.searchElementByPosition(searchPosition);
}
break;
case 10:
listImpl.deleteNodeByGivenData(scanner.nextInt());
break;
case 11:
listImpl.searchElementIteratively(scanner.nextInt());
break;
case 12:
listImpl.searchElementRecursively(listImpl.start, scanner.nextInt(), 0);
break;
default:
System.out.println("invalid choice");
break;
}
} while (yes);
}
}
它会帮助你链接列表。
答案 6 :(得分:3)
链接列表,用于演示Java中的Insert Front,Delete Front,Insert Rear和Delete Rear操作:
import java.io.DataInputStream;
import java.io.IOException;
public class LinkedListTest {
public static void main(String[] args) {
// TODO Auto-generated method stub
Node root = null;
DataInputStream reader = new DataInputStream(System.in);
int op = 0;
while(op != 6){
try {
System.out.println("Enter Option:\n1:Insert Front 2:Delete Front 3:Insert Rear 4:Delete Rear 5:Display List 6:Exit");
//op = reader.nextInt();
op = Integer.parseInt(reader.readLine());
switch (op) {
case 1:
System.out.println("Enter Value: ");
int val = Integer.parseInt(reader.readLine());
root = insertNodeFront(val,root);
display(root);
break;
case 2:
root=removeNodeFront(root);
display(root);
break;
case 3:
System.out.println("Enter Value: ");
val = Integer.parseInt(reader.readLine());
root = insertNodeRear(val,root);
display(root);
break;
case 4:
root=removeNodeRear(root);
display(root);
break;
case 5:
display(root);
break;
default:
System.out.println("Invalid Option");
break;
}
} catch (Exception e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
System.out.println("Exited!!!");
try {
reader.close();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
static Node insertNodeFront(int value, Node root){
Node temp = new Node(value);
if(root==null){
return temp; // as root or first
}
else
{
temp.next = root;
return temp;
}
}
static Node removeNodeFront(Node root){
if(root==null){
System.out.println("List is Empty");
return null;
}
if(root.next==null){
return null; // remove root itself
}
else
{
root=root.next;// make next node as root
return root;
}
}
static Node insertNodeRear(int value, Node root){
Node temp = new Node(value);
Node cur = root;
if(root==null){
return temp; // as root or first
}
else
{
while(cur.next!=null)
{
cur = cur.next;
}
cur.next = temp;
return root;
}
}
static Node removeNodeRear(Node root){
if(root==null){
System.out.println("List is Empty");
return null;
}
Node cur = root;
Node prev = null;
if(root.next==null){
return null; // remove root itself
}
else
{
while(cur.next!=null)
{
prev = cur;
cur = cur.next;
}
prev.next=null;// remove last node
return root;
}
}
static void display(Node root){
System.out.println("Current List:");
if(root==null){
System.out.println("List is Empty");
return;
}
while (root!=null){
System.out.print(root.val+"->");
root=root.next;
}
System.out.println();
}
static class Node{
int val;
Node next;
public Node(int value) {
// TODO Auto-generated constructor stub
val = value;
next = null;
}
}
}
答案 7 :(得分:1)
我是如何创建这样的链接列表的?这是如何工作的? 这是一个链表。带有指向列表中下一个项目的链接的项目。只要您在列表的开头保留对项目的引用,就可以使用每个后续引用来遍历整个事物。
要追加,您需要做的就是找到列表的末尾,并使下一个项目成为您想要追加的值,所以如果下一个项目为非null,则必须在下一个项目上调用append,直到你找到了列表的末尾。
this.next.Append(word);
答案 8 :(得分:0)
请阅读这篇文章:How To Implement a LinkedList Class From Scratch In Java
package com.crunchify.tutorials;
/**
* @author Crunchify.com
*/
public class CrunchifyLinkedListTest {
public static void main(String[] args) {
CrunchifyLinkedList lList = new CrunchifyLinkedList();
// add elements to LinkedList
lList.add("1");
lList.add("2");
lList.add("3");
lList.add("4");
lList.add("5");
/*
* Please note that primitive values can not be added into LinkedList
* directly. They must be converted to their corresponding wrapper
* class.
*/
System.out.println("lList - print linkedlist: " + lList);
System.out.println("lList.size() - print linkedlist size: " + lList.size());
System.out.println("lList.get(3) - get 3rd element: " + lList.get(3));
System.out.println("lList.remove(2) - remove 2nd element: " + lList.remove(2));
System.out.println("lList.get(3) - get 3rd element: " + lList.get(3));
System.out.println("lList.size() - print linkedlist size: " + lList.size());
System.out.println("lList - print linkedlist: " + lList);
}
}
class CrunchifyLinkedList {
// reference to the head node.
private Node head;
private int listCount;
// LinkedList constructor
public CrunchifyLinkedList() {
// this is an empty list, so the reference to the head node
// is set to a new node with no data
head = new Node(null);
listCount = 0;
}
public void add(Object data)
// appends the specified element to the end of this list.
{
Node crunchifyTemp = new Node(data);
Node crunchifyCurrent = head;
// starting at the head node, crawl to the end of the list
while (crunchifyCurrent.getNext() != null) {
crunchifyCurrent = crunchifyCurrent.getNext();
}
// the last node's "next" reference set to our new node
crunchifyCurrent.setNext(crunchifyTemp);
listCount++;// increment the number of elements variable
}
public void add(Object data, int index)
// inserts the specified element at the specified position in this list
{
Node crunchifyTemp = new Node(data);
Node crunchifyCurrent = head;
// crawl to the requested index or the last element in the list,
// whichever comes first
for (int i = 1; i < index && crunchifyCurrent.getNext() != null; i++) {
crunchifyCurrent = crunchifyCurrent.getNext();
}
// set the new node's next-node reference to this node's next-node
// reference
crunchifyTemp.setNext(crunchifyCurrent.getNext());
// now set this node's next-node reference to the new node
crunchifyCurrent.setNext(crunchifyTemp);
listCount++;// increment the number of elements variable
}
public Object get(int index)
// returns the element at the specified position in this list.
{
// index must be 1 or higher
if (index <= 0)
return null;
Node crunchifyCurrent = head.getNext();
for (int i = 1; i < index; i++) {
if (crunchifyCurrent.getNext() == null)
return null;
crunchifyCurrent = crunchifyCurrent.getNext();
}
return crunchifyCurrent.getData();
}
public boolean remove(int index)
// removes the element at the specified position in this list.
{
// if the index is out of range, exit
if (index < 1 || index > size())
return false;
Node crunchifyCurrent = head;
for (int i = 1; i < index; i++) {
if (crunchifyCurrent.getNext() == null)
return false;
crunchifyCurrent = crunchifyCurrent.getNext();
}
crunchifyCurrent.setNext(crunchifyCurrent.getNext().getNext());
listCount--; // decrement the number of elements variable
return true;
}
public int size()
// returns the number of elements in this list.
{
return listCount;
}
public String toString() {
Node crunchifyCurrent = head.getNext();
String output = "";
while (crunchifyCurrent != null) {
output += "[" + crunchifyCurrent.getData().toString() + "]";
crunchifyCurrent = crunchifyCurrent.getNext();
}
return output;
}
private class Node {
// reference to the next node in the chain,
// or null if there isn't one.
Node next;
// data carried by this node.
// could be of any type you need.
Object data;
// Node constructor
public Node(Object dataValue) {
next = null;
data = dataValue;
}
// another Node constructor if we want to
// specify the node to point to.
public Node(Object dataValue, Node nextValue) {
next = nextValue;
data = dataValue;
}
// these methods should be self-explanatory
public Object getData() {
return data;
}
public void setData(Object dataValue) {
data = dataValue;
}
public Node getNext() {
return next;
}
public void setNext(Node nextValue) {
next = nextValue;
}
}
}
<强>输出强>
lList - print linkedlist: [1][2][3][4][5]
lList.size() - print linkedlist size: 5
lList.get(3) - get 3rd element: 3
lList.remove(2) - remove 2nd element: true
lList.get(3) - get 3rd element: 4
lList.size() - print linkedlist size: 4
lList - print linkedlist: [1][3][4][5]
答案 9 :(得分:0)
请找到下面的程序
class Node {
int data;
Node next;
public Node(int data) {
this.data = data;
this.next = null;
}
}
public class LinkedListManual {
Node node;
public void pushElement(int next_node) {
Node nd = new Node(next_node);
nd.next = node;
node = nd;
}
public int getSize() {
Node temp = node;
int count = 0;
while (temp != null) {
count++;
temp = temp.next;
}
return count;
}
public void getElement() {
Node temp = node;
while (temp != null) {
System.out.println(temp.data);
temp = temp.next;
}
}
public static void main(String[] args) {
LinkedListManual obj = new LinkedListManual();
obj.pushElement(1);
obj.pushElement(2);
obj.pushElement(3);
obj.getElement(); //get element
System.out.println(obj.getSize()); //get size of link list
}
}
答案 10 :(得分:-1)
class Node
{
int data;
Node link;
public Node()
{
data=0;
link=null;
}
Node ptr,start,temp;
void create()throws IOException
{
int n;
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
System.out.println("Enter first data");
this.data=Integer.parseInt(br.readLine());
ptr=this;
start=ptr;
char ins ='y';
do
{
System.out.println("Wanna Insert another node???");
ins=(char)br.read();
br.read();
if(ins=='y')
{
temp=new Node();
System.out.println("Enter next data");
temp.data=Integer.parseInt(br.readLine());
temp.link=null;
ptr.link=temp;
temp=null;
ptr=ptr.link;
}
}while(ins=='y');
}
public static void main(String args[])throws IOException
{
Node first= new Node();
first.create();
}
}