算法题（使用递归反转双向链表）

本文概述

C ++
Java
Python3
C#

给定一个双向链表。使用递归将其反转。

Original Doubly linked list

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Reversed Doubly linked list

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我们已经讨论过> 逆向双向链接列表的迭代解决方案
【算法题（使用递归反转双向链表）】算法
1)如果列表为空, 则返回
2)通过交换head-> prev和head-> next来反转头部
3)如果prev = NULL, 则表示列表已完全反转。否则反转(head-> prev)
C ++

//C++ implementation to reverse a doubly //linked list using recursion #include < bits/stdc++.h> using namespace std; //a node of the doubly linked list struct Node { int data; Node *next, *prev; }; //function to get a new node Node* getNode( int data) { //allocate space Node* new_node = new Node; new_node-> data = https://www.lsbin.com/data; new_node-> next = new_node-> prev = NULL; return new_node; }//function to insert a node at the beginning //of the Doubly Linked List void push(Node** head_ref, Node* new_node) { //since we are adding at the beginning, //prev is always NULL new_node-> prev = NULL; //link the old list off the new node new_node-> next = (*head_ref); //change prev of head node to new node if ((*head_ref) != NULL) (*head_ref)-> prev = new_node; //move the head to point to the new node (*head_ref) = new_node; }//function to reverse a doubly linked list Node* Reverse(Node* node) { //If empty list, return if (!node) return NULL; //Otherwise, swap the next and prev Node* temp = node-> next; node-> next = node-> prev; node-> prev = temp; //If the prev is now NULL, the list //has been fully reversed if (!node-> prev) return node; //Otherwise, keep going return Reverse(node-> prev); }//Function to print nodes in a given doubly //linked list void printList(Node* head) { while (head != NULL) { cout < < head-> data < <" " ; head = head-> next; } }//Driver program to test above int main() { //Start with the empty list Node* head = NULL; //Create doubly linked: 10< -> 8< -> 4< -> 2 */ push(& head, getNode(2)); push(& head, getNode(4)); push(& head, getNode(8)); push(& head, getNode(10)); cout < < "Original list: " ; printList(head); //Reverse doubly linked list head = Reverse(head); cout < < "\nReversed list: " ; printList(head); return 0; }

Java

//Java implementation to reverse a doubly //linked list using recursion class GFG {//a node of the doubly linked list static class Node { int data; Node next, prev; }; //function to get a new node static Node getNode( int data) { //allocate space Node new_node = new Node(); new_node.data = https://www.lsbin.com/data; new_node.next = new_node.prev = null ; return new_node; } //function to insert a node at the beginning //of the Doubly Linked List static Node push(Node head_ref, Node new_node) { //since we are adding at the beginning, //prev is always null new_node.prev = null ; //link the old list off the new node new_node.next = (head_ref); //change prev of head node to new node if ((head_ref) != null ) (head_ref).prev = new_node; //move the head to point to the new node (head_ref) = new_node; return head_ref; } //function to reverse a doubly linked list static Node Reverse(Node node) { //If empty list, return if (node == null ) return null ; //Otherwise, swap the next and prev Node temp = node.next; node.next = node.prev; node.prev = temp; //If the prev is now null, the list //has been fully reversed if (node.prev == null ) return node; //Otherwise, keep going return Reverse(node.prev); } //Function to print nodes in a given doubly //linked list static void printList(Node head) { while (head != null ) { System.out.print( head.data +" " ); head = head.next; } } //Driver code public static void main(String args[]) { //Start with the empty list Node head = null ; //Create doubly linked: 10< .8< .4< .2 / head = push(head, getNode( 2 )); head = push(head, getNode( 4 )); head = push(head, getNode( 8 )); head = push(head, getNode( 10 )); System.out.print( "Original list: " ); printList(head); //Reverse doubly linked list head = Reverse(head); System.out.print( "\nReversed list: " ); printList(head); } }//This code is contributed by Arnab Kundu

Python3

# Python3 implementation to reverse a doubly # linked list using recursion import math# a node of the doubly linked list class Node: def __init__( self , data): self .data = https://www.lsbin.com/data self . next = None# function to get a new node def getNode(data):# allocate space new_node = Node(data) new_node.data = data new_node. next = new_node.prev = None return new_node# function to insert a node at the beginning # of the Doubly Linked List def push(head_ref, new_node):# since we are adding at the beginning, # prev is always None new_node.prev = None# link the old list off the new node new_node. next = head_ref# change prev of head node to new node if (head_ref ! = None ): head_ref.prev = new_node# move the head to point to the new node head_ref = new_node return head_ref# function to reverse a doubly linked list def Reverse(node):# If empty list, return if not node: return None# Otherwise, swap the next and prev temp = node. next node. next = node.prev node.prev = temp# If the prev is now None, the list # has been fully reversed if not node.prev: return node# Otherwise, keep going return Reverse(node.prev)# Function to print nodes in a given doubly # linked list def printList(head): while (head ! = None ) : print (head.data, end =" " ) head = head. next# Driver Code if __name__ = = '__main__' : # Start with the empty list head = None# Create doubly linked: 10< .8< .4< .2 */ head = push(head, getNode( 2 )); head = push(head, getNode( 4 )); head = push(head, getNode( 8 )); head = push(head, getNode( 10 )); print ( "Original list: " , end = "") printList(head)# Reverse doubly linked list head = Reverse(head) print ( "\nReversed list: " , end = "") printList(head)# This code is contributed by Srathore

//C# implementation to reverse a doubly using System; //linked list using recursion class GFG { //a node of the doubly linked list public class Node { public int data; public Node next, prev; }; //function to get a new node static Node getNode( int data) { //allocate space Node new_node = new Node(); new_node.data = https://www.lsbin.com/data; new_node.next = new_node.prev = null ; return new_node; } //function to insert a node at the beginning //of the Doubly Linked List static Node push(Node head_ref, Node new_node) { //since we are adding at the beginning, //prev is always null new_node.prev = null ; //link the old list off the new node new_node.next = (head_ref); //change prev of head node to new node if ((head_ref) != null ) (head_ref).prev = new_node; //move the head to point to the new node (head_ref) = new_node; return head_ref; } //function to reverse a doubly linked list static Node Reverse(Node node) { //If empty list, return if (node == null ) return null ; //Otherwise, swap the next and prev Node temp = node.next; node.next = node.prev; node.prev = temp; //If the prev is now null, the list //has been fully reversed if (node.prev == null ) return node; //Otherwise, keep going return Reverse(node.prev); } //Function to print nodes in a given doubly //linked list static void printList(Node head) { while (head != null ) { Console.Write( head.data +" " ); head = head.next; } } //Driver code public static void Main(String []argsS) { //Start with the empty list Node head = null ; //Create doubly linked: 10< .8< .4< .2 / head = push(head, getNode(2)); head = push(head, getNode(4)); head = push(head, getNode(8)); head = push(head, getNode(10)); Console.Write( "Original list: " ); printList(head); //Reverse doubly linked list head = Reverse(head); Console.Write( "\nReversed list: " ); printList(head); } } //This code is contributed by Arnab Kundu

输出如下：