问：从双向链表中间删除新节点的程序。

2025年3月17日 | 阅读11分钟

说明

在此程序中，我们将创建一个双向链表并从列表的中间删除一个节点。如果列表为空，则显示消息“列表为空”。如果列表不为空，我们将计算列表的大小，然后将其除以 2 以获取列表的中间点。当前指针将指向头节点。我们将遍历列表直到到达中间点。现在当前指针将指向中间节点。我们删除中间节点，使其前一个节点指向当前节点的下一个节点。

Program to delete a new node from the middle of the doubly linked list

考虑上面的示例，上面列表的中间点是 3。从头节点到中间点迭代当前指针。现在，当前指针指向需要删除的中间节点。在这种情况下，节点 new 是需要删除的中间节点。可以通过使节点 2（当前节点的上一个节点）指向节点 3（当前节点的下一个节点）来删除 new。将当前指针设置为 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。
为创建双向链表定义另一个类，它有两个节点：head 和 tail。最初，head 和 tail 将指向 null。
deleteFromMid() 将从链表中删除中间节点
1. 它首先检查 head 是否为 null（链表为空），然后，它将从函数返回，因为链表中没有节点。
2. 如果列表不为空，它将检查列表是否只有一个节点。
3. 如果链表包含多个节点，则它将计算链表的大小。将大小除以 2 并将其存储在变量 mid 中。
4. 遍历列表直到当前指针指向列表的中间节点。
5. 将当前节点的上一个节点连接到当前节点的下一个节点。
6. 通过将 Current 设置为 null 来删除 Current 节点。
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 DeleteMid:
    #Represent the head and tail of the doubly linked list
    def __init__(self):
        self.head = None;
        self.tail = None;
        self.size = 0;
        
    #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;
        #Size will count the number of nodes present in the list
        self.size = self.size + 1;
        
    #deleteFromMid() will delete a node from middle of the list
    def deleteFromMid(self):
        #Checks whether list is empty
        if(self.head == None):
            return;
        else:
            #current will point to head
            current = self.head;
            
            #Store the mid position of the list
            mid = (self.size//2) if(self.size % 2 == 0) else((self.size+1)//2);
            
            #Iterate through list till current points to mid position
            for i in range(1, mid):
                current = current.next;
                
            #If middle node is head of the list
            if(current == self.head):
                self.head = current.next;
            #If middle node is tail of the list
            elif(current == self.tail):
                self.tail = self.tail.previous;
            else:
                current.previous.next = current.next;
                current.next.previous = current.previous;
            #Delete the middle node
            current = None;
        self.size = self.size - 1;    
    
    #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;
        while(current != None):
            #Prints each node by incrementing pointer.
            print(current.data),
            current = current.next;
        print();
       
dList = DeleteMid();
#Add nodes to the list
dList.addNode(1);
dList.addNode(2);
dList.addNode(3);
dList.addNode(4);
dList.addNode(5);
 
#Printing original list
print("Original List: ");
dList.display();
while(dList.head != None):
    dList.deleteFromMid();
    #Printing updated list
    print("Updated List: ")
    dList.display();

输出

Original List: 
1 2 3 4 5 
Updated List: 
1 2 4 5 
Updated List: 
1 4 5 
Updated List: 
1 5 
Updated List: 
5 
Updated List: 
List is empty

C

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

struct node{
    int data;
    struct node *previous;
    struct node *next;
};    
 
int size = 0;
//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;
    }
    //Size will count the number of nodes present in the list
    size++;
}
 
//deleteFromMid() will delete a node from middle of the list
void deleteFromMid() {
    //Checks whether list is empty
    if(head == NULL) {
        return;
    }
    else {
        //current will point to head
        struct node *current = head;
        
        //Store the mid position of the list
        int mid = (size % 2 == 0) ? (size/2) : ((size+1)/2);    
        
        //Iterate through list till current points to mid position
        for(int i = 1; i < mid; i++){
            current = current->next;
        }
        
        //If middle node is head of the list
        if(current == head) {
            head = current->next;
        }
        //If middle node is tail of the list
        else if(current == tail) {
            tail = tail->previous;
        }
        else {
            current->previous->next = current->next;
            current->next->previous = current->previous;
        }
        //Delete the middle node
        current = NULL;
     }
    size--;        
}
 
//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;
    }
    while(current != NULL) {
        //Prints each node by incrementing pointer.
        printf("%d ",current->data);
        current = current->next;
    }
    printf("\n");
}
    
int main()
{
 
    //Add nodes to the list
    addNode(1);
    addNode(2);
    addNode(3);
    addNode(4);
    addNode(5);
    
    //Printing original list
    printf("Original List: \n");
    display();
    while(head != NULL) {
        deleteFromMid();
        //Printing updated list
        printf("Updated List: \n");
        display();
    }    
 
    return 0;
}

输出

Original List: 
1 2 3 4 5 
Updated List: 
1 2 4 5 
Updated List: 
1 4 5 
Updated List: 
1 5 
Updated List: 
5 
Updated List: 
List is empty

JAVA

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

    class Node{
        int data;
        Node previous;
        Node next;
        
        public Node(int data) {
            this.data = data;
        }
    }
    
    public int size = 0;
    //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;
        }
        //Size will count the number of nodes present in the list
        size++;
    }
    
    //deleteFromMid() will delete a node from middle of the list
    public void deleteFromMid() {
        //Checks whether list is empty
        if(head == null) {
            return;
        }
        else {
            //current will point to head
            Node current = head;
            
            //Store the mid position of the list
            int mid = (size % 2 == 0) ? (size/2) : ((size+1)/2);    
            
            //Iterate through list till current points to mid position
            for(int i = 1; i < mid; i++){
                current = current.next;
            }
            
            //If middle node is head of the list
            if(current == head) {
                head = current.next;
            }
            //If middle node is tail of the list
            else if(current == tail) {
                tail = tail.previous;
            }
            else {
                current.previous.next = current.next;
                current.next.previous = current.previous;
            }
            //Delete the middle node
            current = null;
        }
        size--;        
    }
    
    //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;
        }
        while(current != null) {
            //Prints each node by incrementing the pointer.

            System.out.print(current.data + " ");
            current = current.next;
        }
        System.out.println();
    }
    
    public static void main(String[] args) {
        
        DeleteMid dList = new DeleteMid();
        //Add nodes to the list
        dList.addNode(1);
        dList.addNode(2);
        dList.addNode(3);
        dList.addNode(4);
        dList.addNode(5);
        
        //Printing original list
        System.out.println("Original List: ");
        dList.display();
        while(dList.head != null) {
            dList.deleteFromMid();
            //Printing updated list
            System.out.println("Updated List: ");
            dList.display();
        }
    }
}

输出

Original List: 
1 2 3 4 5 
Updated List: 
1 2 4 5 
Updated List: 
1 4 5 
Updated List: 
1 5 
Updated List: 
5 
Updated List: 
List is empty

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 DeleteMid<T>{
            public int size = 0;
            //Represent the head and tail of the doubly linked list
            public Node<T> head = null;             
             public 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;
                }
                //Size will count the number of nodes present in the list
                size++;
            }
    
            //deleteFromMid() will delete a node from middle of the list
            public void deleteFromMid() {
                //Checks whether list is empty
                if(head == null) {
                    return;
                }
                else {
                    //current will point to head
                    Node<T> current = head;
 
                    //Store the mid position of the list
                    int mid = (size % 2 == 0) ? (size/2) : ((size+1)/2);    
 
                    //Iterate through list till current points to mid position
                    for(int i = 1; i < mid; i++){
                        current = current.next;
                    }
 
                    //If middle node is head of the list
                    if(current == head) {
                        head = current.next;
                    }
                    //If middle node is tail of the list
                    else if(current == tail) {
                        tail = tail.previous;
                    }
                    else {
                        current.previous.next = current.next;
                        current.next.previous = current.previous;
                    }
                    //Delete the middle node
                    current = null;
                }
                size--;        
            }
    
            //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;
                }
                while(current != null) {
                    //Prints each node by incrementing the pointer.

                    Console.Write(current.data + " ");
                    current = current.next;
                }
                Console.WriteLine();
            }
        }
        
        public static void Main()
        {
            DeleteMid<int> dList = new DeleteMid<int>();
            //Add nodes to the list
            dList.addNode(1);
            dList.addNode(2);
            dList.addNode(3);
            dList.addNode(4);
            dList.addNode(5);
 
            //Printing original list
            Console.WriteLine("Original List: ");
            dList.display();
            while(dList.head != null) {
                dList.deleteFromMid();
                //Printing updated list
                Console.WriteLine("Updated List: ");
                dList.display();
            }
        }    
    }
}

输出

Original List: 
1 2 3 4 5 
Updated List: 
1 2 4 5 
Updated List: 
1 4 5 
Updated List: 
1 5 
Updated List: 
5 
Updated List: 
List is empty

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 DeleteMid{
    //Represent the head and tail of the doubly linked list
    public $head;
    public $tail;
    public $size;
    function __construct(){
        $this->head = NULL;
        $this->tail = NULL;
        $this->size = 0;
    }
    
    //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;
        }
        //Size will count the number of nodes present in the list
        $this->size++;
    }
    
    //deleteFromMid() will delete a node from middle of the list
    function deleteFromMid() {
        //Checks whether list is empty
        if($this->head == NULL) {
            return;
        }
        else {
            //current will point to head
            $current = $this->head;
            
            //Store the mid position of the list
            $mid = ($this->size % 2 == 0) ? ($this->size/2) : (($this->size+1)/2);    
            
            //Iterate through list till current points to mid position
            for($i = 1; $i < $mid; $i++){
                $current = $current->next;
            }
            
            //If middle node is head of the list
            if($current == $this->head) {
                $this->head = $current->next;
            }
            //If middle node is tail of the list
            else if($current == $this->tail) {
                $this->tail = $this->tail->previous;
            }
            else {
                $current->previous->next = $current->next;
                $current->next->previous = $current->previous;
            }
            //Delete the middle node
            $current = NULL;
        }
        $this->size--;        
    }
    
    //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;
        }
        while($current != NULL) {
            //Prints each node by incrementing pointer.
            print($current->data . " ");
            $current = $current->next;
        }
        print("<br>");
    }
}
    
$dList = new DeleteMid();
//Add nodes to the list
$dList->addNode(1);
$dList->addNode(2);
$dList->addNode(3);
$dList->addNode(4);
$dList->addNode(5);
 
//Printing original list
print("Original List: <br>");
$dList->display();
while($dList->head != NULL) {
    $dList->deleteFromMid();
    //Printing updated list
    print("Updated List: <br>");
    $dList->display();
}
?>
</body>
</html>

输出

Original List: 
1 2 3 4 5 
Updated List: 
1 2 4 5 
Updated List: 
1 4 5 
Updated List: 
1 5 
Updated List: 
5 
Updated List: 
List is empty

下一主题#

问：从双向链表中间删除新节点的程序。

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问：从双向链表中间删除新节点的程序。

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