问：创建和显示双向链表的程序。

2025年3月17日 | 阅读 8 分钟

说明

在这个程序中，我们将创建一个双向链表并打印列表中存在的所有节点。

双向链表

双向链表是链表的一种变体。链表是一种线性数据结构，可以描述为节点的集合。节点通过指针连接。每个节点包含两个字段：数据和指向下一个字段的指针。链表的第一个节点称为头节点，链表的最后一个节点称为尾节点。

单向链表的局限性之一是它只能在一个方向（即向前）遍历。双向链表通过提供一个指向前一个节点的额外指针克服了这一限制。借助前一个指针，双向链表可以向后遍历，从而使插入和删除操作更容易执行。因此，双向链表中的典型节点由三个字段组成：

数据表示存储在节点中的数据值。

前一个表示指向前一个节点的指针。

Next 表示指向列表中下一个节点的指针。

Program to create and display a doubly linked list

上图表示了一个双向链表，其中每个节点都有两个指针，分别指向前一个节点和后一个节点。这里，节点1代表链表的头部。头部节点的前驱指针将始终指向NULL。节点1的后继指针将指向节点2。节点5代表链表的尾部，其前驱指针将指向节点4，后继指针将指向NULL。

算法

定义一个 Node 类，它表示列表中的一个节点。它将有三个属性：数据、previous 指向前一个节点，以及 next 指向下一个节点。
定义另一个用于创建双向链表的类，它有两个节点：头节点和尾节点。最初，头节点和尾节点将指向 null。
addNode() 将向列表中添加节点。
1. 它首先检查头节点是否为 null，然后将节点作为头节点插入。
2. 头节点和尾节点都将指向新添加的节点。
3. 头节点的前一个指针将指向 null，尾节点的下一个指针将指向 null。
4. 如果头节点不为 null，则新节点将插入到列表的末尾，使得新节点的前一个指针将指向尾节点。
5. 新节点将成为新的尾节点。尾节点的下一个指针将指向 null。
display() 将显示列表中存在的所有节点。
1. 定义一个新节点“current”，它将指向头节点。
2. 打印 current.data 直到 current 指向 null。
3. 在每次迭代中，current 将指向列表中的下一个节点。

解决方案

Python

#Represent a node of doubly linked list
class Node:
    def __init__(self,data):
        self.data = data;
        self.previous = None;
        self.next = None;
        
class DoublyLinkedList:
    #Represent the head and tail of the doubly linked list
    def __init__(self):
        self.head = None;
        self.tail = None;
        
    #addNode() will add a node to the list
    def addNode(self, data):
        #Create a new node
        newNode = Node(data);
        
        #If list is empty
        if(self.head == None):
            #Both head and tail will point to newNode
            self.head = self.tail = newNode;
            #head's previous will point to None
            self.head.previous = None;
            #tail's next will point to None, as it is the last node of the list
            self.tail.next = None;
        else:
            #newNode will be added after tail such that tail's next will point to newNode
            self.tail.next = newNode;
            #newNode's previous will point to tail
            newNode.previous = self.tail;
            #newNode will become new tail
            self.tail = newNode;
            #As it is last node, tail's next will point to None
            self.tail.next = None;
            
    #display() will print out the nodes of the list
    def display(self):
        #Node current will point to head
        current = self.head;
        if(self.head == None):
            print("List is empty");
            return;
        print("Nodes of doubly linked list: ");
        while(current != None): 
            #Prints each node by incrementing pointer.
            print(current.data),;
            current = current.next;
            
dList = DoublyLinkedList();
#Add nodes to the list
dList.addNode(1);
dList.addNode(2);
dList.addNode(3);
dList.addNode(4);
dList.addNode(5);
 
#Displays the nodes present in the list
dList.display();

输出

Nodes of doubly linked list: 
1 2 3 4 5

C

#include <stdio.h>
 
//Represent a node of the doubly linked list

struct node{
    int data;
    struct node *previous;
    struct node *next;
};    
 
//Represent the head and tail of the doubly linked list
struct node *head, *tail = NULL;
 
//addNode() will add a node to the list
void addNode(int data) {
    //Create a new node
    struct node *newNode = (struct node*)malloc(sizeof(struct node));
    newNode->data = data;
    
    //If list is empty
    if(head == NULL) {
        //Both head and tail will point to newNode
        head = tail = newNode;
        //head's previous will point to NULL
        head->previous = NULL;
        //tail's next will point to NULL, as it is the last node of the list
        tail->next = NULL;
    }
    else {
        //newNode will be added after tail such that tail's next will point to newNode
        tail->next = newNode;
        //newNode's previous will point to tail
        newNode->previous = tail;
        //newNode will become new tail
        tail = newNode;
        //As it is last node, tail's next will point to NULL
        tail->next = NULL;
    }
}
 
//display() will print out the nodes of the list
void display() {
    //Node current will point to head
    struct node *current = head;
    if(head == NULL) {
        printf("List is empty\n");
        return;
    }
    printf("Nodes of doubly linked list: \n");
    while(current != NULL) {
        //Prints each node by incrementing pointer.
        printf("%d ", current->data);
        current = current->next;
    }
}
 
int main()
{
    //Add nodes to the list
    addNode(1);
    addNode(2);
    addNode(3);
    addNode(4);
    addNode(5);
    
    //Displays the nodes present in the list
    display();
 
    return 0;
	}

输出

Nodes of doubly linked list: 
1 2 3 4 5

JAVA

public class DoublyLinkedList {
    
    //Represent a node of the doubly linked list

    class Node{
        int data;
        Node previous;
        Node next;
        
        public Node(int data) {
            this.data = data;
        }
    }
    
    //Represent the head and tail of the doubly linked list
    Node head, tail = null;
    
    //addNode() will add a node to the list
    public void addNode(int data) {
        //Create a new node
        Node newNode = new Node(data);
        
        //If list is empty
        if(head == null) {
            //Both head and tail will point to newNode
            head = tail = newNode;
            //head's previous will point to null
            head.previous = null;
            //tail's next will point to null, as it is the last node of the list
            tail.next = null;
        }
        else {
            //newNode will be added after tail such that tail's next will point to newNode
            tail.next = newNode;
            //newNode's previous will point to tail
            newNode.previous = tail;
            //newNode will become new tail
            tail = newNode;
            //As it is last node, tail's next will point to null
            tail.next = null;
        }
    }
    
    //display() will print out the nodes of the list
    public void display() {
        //Node current will point to head
        Node current = head;
        if(head == null) {
            System.out.println("List is empty");
            return;
        }
        System.out.println("Nodes of doubly linked list: ");
        while(current != null) {
            //Prints each node by incrementing the pointer.

            System.out.print(current.data + " ");
            current = current.next;
        }
    }
    
    public static void main(String[] args) {
        
        DoublyLinkedList dList = new DoublyLinkedList();
        //Add nodes to the list
        dList.addNode(1);
        dList.addNode(2);
        dList.addNode(3);
        dList.addNode(4);
        dList.addNode(5);
        
        //Displays the nodes present in the list
        dList.display();
    }
}

输出

Nodes of doubly linked list: 
1 2 3 4 5

C#

using System; 
namespace DoublyLinkedList 
{                     
    public class Program
    {
        //Represent a node of the doubly linked list

        public class Node<T>{
            public T data;
            public Node<T> previous;
            public Node<T> next;
            
            public Node(T value) {
                data = value;
            }
        }
        
        public class DoublyLinkedList<T>{
            //Represent the head and tail of the doubly linked list
            protected Node<T> head = null;             
             protected Node<T> tail = null;
            
            //addNode() will add a node to the list
            public void addNode(T data) {
                //Create a new node
                Node<T> newNode = new Node<T>(data);
 
                //If list is empty
                if(head == null) {
                    //Both head and tail will point to newNode
                    head = tail = newNode;
                    //head's previous will point to null
                    head.previous = null;
                    //tail's next will point to null, as it is the last node of the list
                    tail.next = null;
                }
                else {
                    //newNode will be added after tail such that tail's next will point to newNode
                    tail.next = newNode;
                    //newNode's previous will point to tail
                    newNode.previous = tail;
                    //newNode will become new tail
                    tail = newNode;
                    //As it is last node, tail's next will point to null
                    tail.next = null;
                }
            }
    
            //display() will print out the nodes of the list
            public void display() {
                //Node current will point to head
                Node<T> current = head;
                if(head == null) {
                    Console.WriteLine("List is empty");
                    return;
                }
                Console.WriteLine("Nodes of doubly linked list: ");
                while(current != null) {
                    //Prints each node by incrementing the pointer.

                    Console.Write(current.data + " ");
                    current = current.next;
                }
            }
        }
        
        public static void Main()
        {
            DoublyLinkedList<int> dList = new DoublyLinkedList<int>();
            //Add nodes to the list
            dList.addNode(1);
            dList.addNode(2);
            dList.addNode(3);
            dList.addNode(4);
            dList.addNode(5);
 
            //Displays the nodes present in the list
            dList.display();
        }    
    }
}

输出

Nodes of doubly linked list: 
1 2 3 4 5

PHP

 <!DOCTYPE html>
<html>
<body>
<?php
//Represent a node of doubly linked list
class Node{
    public $data;
    public $previous;
    public $next;
    
    function __construct($data){
        $this->data = $data;
    }
}
class DoublyLinkedList{
    //Represent the head and tail of the doubly linked list
    public $head;
    public $tail;
    function __construct(){
        $this->head = NULL;
        $this->tail = NULL;
    }
    
    //addNode() will add a node to the list
    function addNode($data){
        //Create a new node
        $newNode = new Node($data);
        
        //If list is empty
        if($this->head == NULL) {
            //Both head and tail will point to newNode
            $this->head = $this->tail = $newNode;
            //head's previous will point to NULL
            $this->head->previous = NULL;
            //tail's next will point to NULL, as it is the last node of the list
            $this->tail->next = NULL;
        }
        else {
            //newNode will be added after tail such that tail's next will point to newNode
            $this->tail->next = $newNode;
            //newNode's previous will point to tail
            $newNode->previous = $this->tail;
            //newNode will become new tail
            $this->tail = $newNode;
            //As it is last node, tail's next will point to NULL
            $this->tail->next = NULL;
        }
    }
    
    //display() will print out the nodes of the list
    function display() {
        //Node current will point to head
        $current = $this->head;
        if($this->head == NULL) {
            print("List is empty <br>");
            return;
        }
        print("Nodes of doubly linked list: <br>");
        while($current != NULL) {
            //Prints each node by incrementing pointer.
            print($current->data . " ");
            $current = $current->next;
        }
    }
}
    
$dList = new DoublyLinkedList();
//Add nodes to the list
$dList->addNode(1);
$dList->addNode(2);
$dList->addNode(3);
$dList->addNode(4);
$dList->addNode(5);
 
//Displays the nodes present in the list
$dList->display();
?>
</body>
</html>

输出

Nodes of doubly linked list: 
1 2 3 4 5

下一主题#

问：创建和显示双向链表的程序。

说明

算法

解决方案

Python

C

JAVA

C#

PHP

联系信息

关注我们

教程

面试题

在线编译器

Python

Java

.Net Framework

AI, ML and Data Science

Cloud Technology

B.Tech and MCA

Web Technology

PHP

Software Testing

Technical Interview

Java Interview

Python

Web Interview

Database Interview

B.Tech / MCA

Important Interview

Software Testing Interview

Company Interviews

Online Compilers

Multiple Choice Questions

程序

问：创建和显示双向链表的程序。

说明

算法

解决方案

Python

C

JAVA

C#

PHP

相关帖子

查找二叉树最大宽度的程序

将千米转换为厘米的程序

在单链表中搜索元素的程序

计算球体体积的程序

打印字母图案 11 的程序

确定两个字符串是否为变位词（anagram）的程序

查找字符串的所有可能子集的程序

打印数组所有元素之和的程序

编写查找四面体（Tetrahedron）面积的程序

打印 1 到 100 之间的所有丰数的程序

订阅 Tpoint Tech

联系信息

关注我们

教程

面试题

在线编译器