Last Updated on March 30, 2022 by Ria Pathak
Concepts Used
Linked list,recursion.
Difficulty Level
Easy.
Problem Statement :
Given pointer to the head node of a linked list, the task is to reverse the linked list.
See original problem statement here
Example:
Assume that we have linked list 1 → 2 → 3 → Ø,
we have to change it to Ø ← 1 ← 2 ← 3.Explanation:
Approach 1(Iterative approach):
While you are traversing the list, change the current node’s next pointer to point to its previous element. Since a node does not have reference to its previous node, you must store its previous element beforehand. You also need another pointer to store the next node before changing the reference. Do not forget to return the new head reference at the end!
struct Node* Reverse(Node *head)
{
Node *tail, *t;
tail = NULL;
while (head != NULL)
{
t = head->next;
head->next = tail;
tail = head;
head = t;
}
return tail;
// Complete this method
}
public static Node reverseLinkedList(Node currentNode)
{
// For first node, previousNode will be null
Node previousNode=null;
Node nextNode;
while(currentNode!=null)
{
nextNode=currentNode.next;
// reversing the link
currentNode.next=previousNode;
// moving currentNode and previousNode by 1 node
previousNode=currentNode;
currentNode=nextNode;
}
return previousNode;
}
def Reverse(head):
tail = NULL
while (head != NULL):
t = head.next
head.next = tail
tail = head
head = t
return tail
Approach 2(Using recursion):
We can find it easier to start from the bottom up,by asking and answering tiny questions:
What is the reverse of NULL? NULL.
What is the reverse of a one element list?The element itself.
What is the reverse of an n element list? The reverse of the second element followed by the first element.
Refer to commented implementation
Node Reverse(Node head) {
/* We have two conditions in this if statement.
This first condition immediately returns null
when the list is null. The second condition returns
the final node in the list. That final node is sent
into the "remaining" Node below.
-----------------------------------------------------*/
if (head == null || head.next == null) {
return head;
}
/* When the recursion creates the stack for A -> B -> C
(RevA(RevB(RevC()))) it will stop at the last node and
the recursion will end, beginning the unraveling of the
nested functions from the inside, out.
-----------------------------------------------------*/
Node remaining = Reverse(head.next);
/* Now we have the "remaining" node returned and accessible
to the node prior. This remaining node will be returned
by each function as the recursive stack unravels.
Assigning head to head.next.next where A is the head
and B is after A, (A -> B), would set B's pointer to A,
reversing their direction to be A <- B.
-----------------------------------------------------*/
head.next.next = head;
/* Now that those two elements are reversed, we need to set
the pointer of the new tail-node to null.
-----------------------------------------------------*/
head.next = null;
/* Now we return remaining so that remaining is always
reassigned to itself and is eventually returned by the
first function call.
-----------------------------------------------------*/
return remaining;
}SOLUTION:
#include <stdio.h>
#include <stdlib.h>
/* A structure of linked list node */
struct node {
int data;
struct node *next;
} *head;
void initialize(){
head = NULL;
}
/*
Given a Inserts a node in front of a singly linked list.
*/
void insert(int num) {
/* Create a new Linked List node */
struct node* newNode = (struct node*) malloc(sizeof(struct node));
newNode->data = num;
/* Next pointer of new node will point to head node of linked list */
newNode->next = head;
/* make new node as new head of linked list */
head = newNode;
printf("Inserted Element : %d\n", num);
}
/* Reverses a Linked List using recursion */
void reverse(struct node* nodePtr) {
/* empty list */
if (nodePtr == NULL)
return;
/* Last node (tail node)*/
if(nodePtr->next == NULL){
head = nodePtr;
return;
}
/* reverse the rest of list and put the first element at the end */
reverse(nodePtr->next);
nodePtr->next->next = nodePtr;
nodePtr->next = NULL;
}
/*
Prints a linked list from head node till tail node
*/
void printLinkedList(struct node *nodePtr) {
while (nodePtr != NULL) {
printf("%d", nodePtr->data);
nodePtr = nodePtr->next;
if(nodePtr != NULL)
printf("-->");
}
}
int main() {
initialize();
/* Creating a linked List*/
insert(1);
insert(2);
insert(3);
insert(4);
insert(5);
printf("\nLinked List\n");
printLinkedList(head);
/* Reverse Linked List */
reverse(head);
printf("\nReversed Linked List\n");
printLinkedList(head);
return 0;
}
#include <bits stdc++.h="">
using namespace std;
struct node{
int data;
node* next;
};
node* newnode(int x)
{
node* temp=new node();
temp->data=x;
temp->next=NULL;
return temp;
}
node* reverse(node* head)
{
if(head==NULL||head->next==NULL)return head;
node* restlist=reverse(head->next);
head->next->next=head;
head->next=NULL;
return restlist;
}
int main()
{
//write your code here
int t;cin>>t;
while(t--)
{
int n;
cin>>n;
int x;cin>>x;
node* temp=newnode(x);
node* head=temp;
for(int i=1;i<n;i++) {="" int="" x;cin="">>x;
head->next=newnode(x);
head=head->next;
}
node* head1=reverse(temp);
while(head1!=NULL)
{
cout<<head1->data<<" ";
head1=head1->next;
}
cout<<"\n";
}
return 0;
}
import java.io.*;
import java.util.*;
public class Main{
static class SinglyLinkedListNode {
public int data;
public SinglyLinkedListNode next;
public SinglyLinkedListNode(int nodeData) {
this.data = nodeData;
this.next = null;
}
}
static class SinglyLinkedList {
public SinglyLinkedListNode head;
public SinglyLinkedListNode tail;
public SinglyLinkedList() {
this.head = null;
this.tail = null;
}
public void insertNode(int nodeData) {
SinglyLinkedListNode node = new SinglyLinkedListNode(nodeData);
if (this.head == null) {
this.head = node;
} else {
this.tail.next = node;
}
this.tail = node;
}
}
static void printLinkedList(SinglyLinkedListNode head)
{
SinglyLinkedListNode temp=head;
while(temp!=null)
{
System.out.print(temp.data+" ");
temp=temp.next;
}
System.out.println();
}
static SinglyLinkedListNode reverseLinkedList(SinglyLinkedListNode head) {
SinglyLinkedListNode prev = null;
SinglyLinkedListNode current = head;
SinglyLinkedListNode next = null;
while (current != null) {
next = current.next;
current.next = prev;
prev = current;
current = next;
}
head = prev;
return head;
}
private static final Scanner scanner = new Scanner(System.in);
public static void main(String[] args) throws IOException {
int testCases = scanner.nextInt();
while (testCases-- > 0) {
SinglyLinkedList llist = new SinglyLinkedList();
int llistCount = scanner.nextInt();
for (int i = 0; i < llistCount; i++) {
int llistItem = scanner.nextInt();
llist.insertNode(llistItem);
}
printLinkedList(reverseLinkedList(llist.head));
}
scanner.close();
}
}
class node:
def __init__(self):
self.data = None
self.next = None
def newnode(x):
temp= node()
temp.data = x
temp.next = None
return temp
def reverse(head):
if(head == None or head.next == None):
return head
restlist = reverse(head.next)
head.next.next = head
head.next = None
return restlist
for _ in range(int(input())):
n = int(input())
a = list(map(int,input().split()))
temp = newnode(a[0])
head = temp
for i in range(1, n):
head.next = newnode(a[i])
head = head.next
head1 = reverse(temp)
while head1:
print(head1.data, end = " ")
head1 = head1.next
print()
[forminator_quiz id="1506"]
This article tried to discuss the concept of Linked List, Recursion. Hope this blog helps you understand and solve the problem. To practice more problems on Linked List, Recursion you can check out MYCODE | Competitive Programming.