Skip to content
Related Articles

Related Articles

Program to unfold a folded linked list

View Discussion
Improve Article
Save Article
  • Difficulty Level : Hard
  • Last Updated : 08 Jul, 2022

A linked list L0 -> L1 -> L2 -> ….. -> LN can be folded as L0 -> LN -> L1 -> LN – 1 -> L2 -> ….. 
Given a folded linked list, the task is to unfold and print the original linked list

Examples:  

Input: 1 -> 6 -> 2 -> 5 -> 3 -> 4 
Output: 1 2 3 4 5 6

Input: 1 -> 5 -> 2 -> 4 -> 3 
Output: 1 2 3 4 5 

Approach: Make a recursive call and store the next node in temp pointer, first node will act as head node and the node which is stored in temp pointer will act as a tail of the list. On returning after reaching the base condition link the head and tail to previous head and tail respectively. 

Base condition: If number of nodes is even then the second last node is head and the last node is tail and if the number of nodes is odd then last node will act as head as well as tail.

Below is the implementation of the above approach:  

C++




// C++ implementation of the approach
#include<bits/stdc++.h>
using namespace std;
 
// Node Class
struct Node
{
    int data;
    Node *next;
};
 
// Head of the list
Node *head;
 
// Tail of the list
Node *tail;
 
// Function to print the list
void display()
{
    if (head == NULL)
        return;
         
    Node* temp = head;
 
    while (temp != NULL)
    {
        cout << temp->data << " ";
        temp = temp->next;
    }
    cout << endl;
}
 
// Function to add node in the list
void push(int data)
{
     
    // Create new node
    Node* nn = new Node();
    nn->data = data;
    nn->next = NULL;
 
    // Linking at first position
    if (head == NULL)
    {
        head = nn;
    }
    else
    {
        Node* temp = head;
 
        while (temp->next != NULL)
        {
            temp = temp->next;
        }
 
        // Linking at last in list
        temp->next = nn;
    }
}
 
// Function to unfold the given link list
void unfold(Node* node)
{
    if (node == NULL)
        return;
 
    // This condition will reach if
    // the number of nodes is odd
    // head and tail is same i->e-> last node
    if (node->next == NULL)
    {
        head = tail = node;
        return;
    }
 
    // This base condition will reach if
    // the number of nodes is even
    // mark head to the second last node
    // and tail to the last node
    else if (node->next->next == NULL)
    {
        head = node;
        tail = node->next;
        return;
    }
 
    // Storing next node in temp pointer
    // before making the recursive call
    Node* temp = node->next;
 
    // Recursive call
    unfold(node->next->next);
 
    // Connecting first node to head
    // and mark it as a new head
    node->next = head;
    head = node;
 
    // Connecting tail to second node (temp)
    // and mark it as a new tail
    tail->next = temp;
    tail = temp;
    tail->next = NULL;
}
 
// Driver code
int main()
{
     
    // Adding nodes to the list
    push(1);
    push(5);
    push(2);
    push(4);
    push(3);
 
    // Displaying the original nodes
    display();
 
    // Calling unfold function
    unfold(head);
 
    // Displaying the list
    // after modification
    display();
}
 
// This code is contributed by pratham76

Java




// Java implementation of the approach
public class GFG {
 
    // Node Class
    private class Node {
        int data;
        Node next;
    }
 
    // Head of the list
    private Node head;
 
    // Tail of the list
    private Node tail;
 
    // Function to print the list
    public void display()
    {
 
        if (head == null)
            return;
        Node temp = head;
 
        while (temp != null) {
            System.out.print(temp.data + " ");
            temp = temp.next;
        }
        System.out.println();
    }
 
    // Function to add node in the list
    public void push(int data)
    {
 
        // Create new node
        Node nn = new Node();
        nn.data = data;
        nn.next = null;
 
        // Linking at first position
        if (head == null) {
            head = nn;
        }
        else {
            Node temp = head;
 
            while (temp.next != null) {
                temp = temp.next;
            }
 
            // Linking at last in list
            temp.next = nn;
        }
    }
 
    // Function to unfold the given link list
    private void unfold(Node node)
    {
        if (node == null)
            return;
 
        // This condition will reach if
        // the number of nodes is odd
        // head and tail is same i.e. last node
        if (node.next == null) {
            head = tail = node;
            return;
        }
 
        // This base condition will reach if
        // the number of nodes is even
        // mark head to the second last node
        // and tail to the last node
        else if (node.next.next == null) {
            head = node;
            tail = node.next;
            return;
        }
 
        // Storing next node in temp pointer
        // before making the recursive call
        Node temp = node.next;
 
        // Recursive call
        unfold(node.next.next);
 
        // Connecting first node to head
        // and mark it as a new head
        node.next = head;
        head = node;
 
        // Connecting tail to second node (temp)
        // and mark it as a new tail
        tail.next = temp;
        tail = temp;
        tail.next = null;
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        GFG l = new GFG();
 
        // Adding nodes to the list
        l.push(1);
        l.push(5);
        l.push(2);
        l.push(4);
        l.push(3);
 
        // Displaying the original nodes
        l.display();
 
        // Calling unfold function
        l.unfold(l.head);
 
        // Displaying the list
        // after modification
        l.display();
    }
}

Python3




# Python3 implementation of the approach
 
# Node Class
class Node:
     
    def __init__(self, data):
         
        self.data = data
        self.next = None
 
# Head of the list
head = None
 
# Tail of the list
tail = None
 
# Function to print the list
def display():
     
    if (head == None):
        return
     
    temp = head
 
    while (temp != None):
        print(temp.data, end = " ")
        temp = temp.next
     
    print()
 
# Function to add node in the list
def push(data):
     
    global head, tail
     
    # Create new node
    nn = Node(data)
 
    # Linking at first position
    if (head == None):
        head = nn
    else:
        temp = head
 
        while (temp.next != None):
            temp = temp.next
             
        # Linking at last in list
        temp.next = nn
 
# Function to unfold the given link list
def unfold(node):
     
    global head, tail
     
    if (node == None):
        return
 
    # This condition will reach if
    # the number of nodes is odd
    # head and tail is same i.e. last node
    if (node.next == None):
        head = tail = node
        return
     
    # This base condition will reach if
    # the number of nodes is even
    # mark head to the second last node
    # and tail to the last node
    elif (node.next.next == None):
        head = node
        tail = node.next
        return
     
    # Storing next node in temp pointer
    # before making the recursive call
    temp = node.next
 
    # Recursive call
    unfold(node.next.next)
 
    # Connecting first node to head
    # and mark it as a new head
    node.next = head
    head = node
 
    # Connecting tail to second node (temp)
    # and mark it as a new tail
    tail.next = temp
    tail = temp
    tail.next = None
 
# Driver code
if __name__=='__main__':
 
    # Adding nodes to the list
    push(1)
    push(5)
    push(2)
    push(4)
    push(3)
 
    # Displaying the original nodes
    display()
 
    # Calling unfold function
    unfold(head)
 
    # Displaying the list
    # after modification
    display()
 
# This code is contributed by rutvik_56

C#




// C# implementation of the approach
using System;
public class GFG {
 
    // Node Class
    private class Node {
        public int data;
        public Node next;
    }
 
    // Head of the list
    private Node head;
 
    // Tail of the list
    private Node tail;
 
    // Function to print the list
    public void display()
    {
 
        if (head == null)
            return;
        Node temp = head;
 
        while (temp != null) {
            Console.Write(temp.data + " ");
            temp = temp.next;
        }
        Console.WriteLine();
    }
 
    // Function to add node in the list
    public void push(int data)
    {
 
        // Create new node
        Node nn = new Node();
        nn.data = data;
        nn.next = null;
 
        // Linking at first position
        if (head == null) {
            head = nn;
        }
        else {
            Node temp = head;
 
            while (temp.next != null) {
                temp = temp.next;
            }
 
            // Linking at last in list
            temp.next = nn;
        }
    }
 
    // Function to unfold the given link list
    private void unfold(Node node)
    {
        if (node == null)
            return;
 
        // This condition will reach if
        // the number of nodes is odd
        // head and tail is same i.e. last node
        if (node.next == null) {
            head = tail = node;
            return;
        }
 
        // This base condition will reach if
        // the number of nodes is even
        // mark head to the second last node
        // and tail to the last node
        else if (node.next.next == null) {
            head = node;
            tail = node.next;
            return;
        }
 
        // Storing next node in temp pointer
        // before making the recursive call
        Node temp = node.next;
 
        // Recursive call
        unfold(node.next.next);
 
        // Connecting first node to head
        // and mark it as a new head
        node.next = head;
        head = node;
 
        // Connecting tail to second node (temp)
        // and mark it as a new tail
        tail.next = temp;
        tail = temp;
        tail.next = null;
    }
 
    // Driver code
    public static void Main()
    {
 
        GFG l = new GFG();
 
        // Adding nodes to the list
        l.push(1);
        l.push(5);
        l.push(2);
        l.push(4);
        l.push(3);
 
        // Displaying the original nodes
        l.display();
 
        // Calling unfold function
        l.unfold(l.head);
 
        // Displaying the list
        // after modification
        l.display();
    }
}
/* This code contributed by PrinciRaj1992 */

Javascript




<script>
 
// Javascript implementation of the approach
 
// Represents node of the linked list
class Node {
        constructor() {
                this.data = 0;
                this.next = null;
             }
        }
         
         
// Head of the list
var head = null;
 
// Tail of the list
var tail = null;
 
// Function to print the list
function display()
    {
 
    if (head == null)
        return;
    var temp = head;
 
    while (temp != null) {
        document.write(temp.data + " ");
        temp = temp.next;
    }
        document.write("</br>");
    }
 
// Function to add node in the list
function push( data)
{
 
    // Create new node
    var nn = new Node();
    nn.data = data;
    nn.next = null;
 
    // Linking at first position
    if (head == null) {
        head = nn;
    }
    else {
        var temp = head;
 
        while (temp.next != null) {
            temp = temp.next;
        }
 
        // Linking at last in list
        temp.next = nn;
    }
}
 
// Function to unfold the given link list
function unfold( node)
{
    if (node == null)
        return;
 
    // This condition will reach if
    // the number of nodes is odd
    // head and tail is same i.e. last node
    if (node.next == null) {
        head = tail = node;
        return;
    }
 
    // This base condition will reach if
    // the number of nodes is even
    // mark head to the second last node
    // and tail to the last node
    else if (node.next.next == null) {
        head = node;
        tail = node.next;
        return;
    }
 
    // Storing next node in temp pointer
    // before making the recursive call
    var temp = node.next;
 
    // Recursive call
    unfold(node.next.next);
 
    // Connecting first node to head
    // and mark it as a new head
    node.next = head;
    head = node;
 
    // Connecting tail to second node (temp)
    // and mark it as a new tail
    tail.next = temp;
    tail = temp;
    tail.next = null;
}
 
 
// Driver Code
 
// Adding nodes to the list
push(1);
push(5);
push(2);
push(4);
push(3);
 
// Displaying the original nodes
display();
 
// Calling unfold function
unfold(head);
 
// Displaying the list
// after modification
display();
 
// This code is contributed by jana_sayantan.
</script>

Output

1 5 2 4 3 
1 2 3 4 5 

Time Complexity: O(N), where N is the total number of nodes in the linked list. 
Auxiliary Space: O(N)

Iterative Approach:-

Approach: First we have to segregate the linked list on the basis of even-odd index. Then we reverse the odd part of segregated list and joined with the first list. This approach does not use recursive space.

Below is the implementation of the above approach:  

C++




// C++ implementation of the approach
#include <iostream>
using namespace std;
 
class ListNode {
public:
    int val = 0;
    ListNode* next = nullptr;
 
    ListNode(int val) { this->val = val; }
};
 
ListNode* reverse(ListNode* head)
{
    ListNode *prev = NULL, *temp = head, *copy = NULL;
    while (temp != NULL) {
        copy = temp->next;
        temp->next = prev;
        prev = temp;
        temp = copy;
    }
    return prev;
}
 
void unfold(ListNode* head)
{
    // for segregating the original linked list into two
    // linked list on the basis of even and odd
    int i = 0;
 
    // For storing the previous node
    // and head node for each linked list
    ListNode *prev1 = NULL, *prev2 = NULL, *h1 = head,
             *h2 = head;
 
    while (head != NULL) {
        if (i % 2 == 0) {
            if (prev1 == NULL) {
                h1 = head;
                prev1 = head;
            }
            else {
                prev1->next = head;
                prev1 = head;
            }
        }
        else {
            if (prev2 == NULL) {
                h2 = head;
                prev2 = head;
            }
            else {
                prev2->next = head;
                prev2 = head;
            }
        }
        i++;
        head = head->next;
    }
 
    prev2->next = NULL;
    ListNode* rev
        = reverse(h2); // reverse the second linked list
    prev1->next = rev; // join the first ll with second one
}
 
void printList(ListNode* node)
{
    ListNode* curr = node;
    while (curr != nullptr) {
        cout << curr->val << " ";
        curr = curr->next;
    }
    cout << endl;
}
 
 
int main()
{
    int n;
    ListNode* dummy = new ListNode(-1);
    ListNode* prev = dummy;
    n=5;int i=0;
    int arr[]={1, 5, 2, 4, 3}; //Elements in the linkedlist
    while (i < n) {
        prev->next = new ListNode(arr[i]);
        prev = prev->next;
        i++;
    }
 
    ListNode* head = dummy->next;
   
      printList(head);
    unfold(head);
    printList(head);
 
    return 0;
}
// This code is contributed by Ankit

Java




// Java implementation of the approach
 
class GFG {
 
    static class ListNode {
        int val = 0;
        ListNode next = null;
 
        ListNode(int val) { this.val = val; }
    }
 
    static ListNode reverse(ListNode head)
    {
        ListNode prev = null, temp = head, copy = null;
        while (temp != null) {
            copy = temp.next;
            temp.next = prev;
            prev = temp;
            temp = copy;
        }
        return prev;
    }
 
    static void unfold(ListNode head)
    {
        // for segregating the original linked list into two
        // linked list on the basis of even and odd
        int i = 0;
 
        // For storing the previous node
        // and head node for each linked list
        ListNode prev1 = null, prev2 = null, h1 = head,
                 h2 = head;
 
        while (head != null) {
            if (i % 2 == 0) {
                if (prev1 == null) {
                    h1 = head;
                    prev1 = head;
                }
                else {
                    prev1.next = head;
                    prev1 = head;
                }
            }
            else {
                if (prev2 == null) {
                    h2 = head;
                    prev2 = head;
                }
                else {
                    prev2.next = head;
                    prev2 = head;
                }
            }
            i++;
            head = head.next;
        }
 
        prev2.next = null;
        ListNode rev
            = reverse(h2); // reverse the second linked list
        prev1.next
            = rev; // join the first ll with second one
    }
 
    static void printList(ListNode node)
    {
        ListNode curr = node;
        while (curr != null) {
            System.out.print(curr.val + " ");
            curr = curr.next;
        }
        System.out.println();
    }
 
    public static void main(String[] args)
    {
        int n;
        ListNode dummy = new ListNode(-1);
        ListNode prev = dummy;
        n = 5;
        int i = 0;
        int arr[] = { 1, 5, 2, 4,
                      3 }; // Elements in the linkedlist
        while (i < n) {
            prev.next = new ListNode(arr[i]);
            prev = prev.next;
            i++;
        }
 
        ListNode head = dummy.next;
 
        printList(head);
        unfold(head);
        printList(head);
    }
}
// This code is contributed by Lovely Jain

Python3




# Python code to implement the above approach
class ListNode:
 
    def __init__(self,val):
        self.next = None
        self.val = val
 
def reverse(head):
 
    prev,temp,copy = None,head,None
    while (temp != None):
        copy = temp.next
        temp.next = prev
        prev = temp
        temp = copy
     
    return prev
 
def unfold(head):
 
    # for segregating the original linked list into two
    # linked list on the basis of even and odd
    i = 0
 
    # For storing the previous node
    # and head node for each linked list
    prev1,prev2,h1,h2 = None,None,head,head
 
    while (head != None):
        if (i % 2 == 0):
            if (prev1 == None):
                h1 = head
                prev1 = head
     
            else :
                prev1.next = head
                prev1 = head
             
        else:
            if (prev2 == None):
                h2 = head
                prev2 = head
             
            else:
                prev2.next = head
                prev2 = head
     
        i += 1
        head = head.next
 
 
    prev2.next = None
    rev = reverse(h2) # reverse the second linked list
    prev1.next = rev # join the first ll with second one
 
def printList(node):
 
    curr = node
    while (curr != None):
        print(curr.val,end = " ")
        curr = curr.next
     
    print()
 
# driver code
dummy = ListNode(-1)
prev = dummy
n=5
i=0
arr = [1, 5, 2, 4, 3] #Elements in the linkedlist
 
while (i < n):
    prev.next = ListNode(arr[i])
    prev = prev.next
    i += 1
 
head = dummy.next
 
printList(head)
unfold(head)
printList(head)
 
# this code is contributed by shinjanpatra

C#




// C# implementation of the approach
using System;
 
class GFG {
 
  class ListNode {
    public int val = 0;
    public ListNode next = null;
    public ListNode(int val) { this.val = val; }
  }
 
  static ListNode reverse(ListNode head)
  {
    ListNode prev = null, temp = head, copy = null;
    while (temp != null) {
      copy = temp.next;
      temp.next = prev;
      prev = temp;
      temp = copy;
    }
    return prev;
  }
 
  static void unfold(ListNode head)
  {
 
    // for segregating the original linked list into two
    // linked list on the basis of even and odd
    int i = 0;
 
    // For storing the previous node and head node for
    // each linked list.
    ListNode prev1 = null, prev2 = null, h2 = head;
 
    while (head != null) {
      if (i % 2 == 0) {
        if (prev1 == null) {
          prev1 = head;
        }
        else {
          prev1.next = head;
          prev1 = head;
        }
      }
      else {
        if (prev2 == null) {
          h2 = head;
          prev2 = head;
        }
        else {
          prev2.next = head;
          prev2 = head;
        }
      }
      i++;
      head = head.next;
    }
 
    prev2.next = null;
    ListNode rev = reverse(
      h2); // reverse the second linked list.
    prev1.next = rev; // join the first linked list with
    // second one.
  }
 
  static void printList(ListNode node)
  {
    ListNode curr = node;
    while (curr != null) {
      Console.Write(curr.val + " ");
      curr = curr.next;
    }
    Console.WriteLine();
  }
 
  static public void Main()
  {
 
    // Code
 
    int n;
    ListNode dummy = new ListNode(-1);
    ListNode prev = dummy;
 
    n = 5;
    int i = 0;
    int[] arr = { 1, 5, 2, 4,
                 3 }; // Elements in the linked list.
    while (i < n) {
      prev.next = new ListNode(arr[i]);
      prev = prev.next;
      i++;
    }
    ListNode head = dummy.next;
 
    printList(head);
    unfold(head);
    printList(head);
  }
}
 
// This code is contributed by lokesh (lokeshmvs21).

Javascript




<script>
 
// JavaScript code to implement the above approach
class ListNode {
 
    constructor(val){
        this.next = null;
        this.val = val;
    }
 
}
 
function reverse(head)
{
    let prev = null, temp = head, copy = null;
    while (temp != null) {
        copy = temp.next;
        temp.next = prev;
        prev = temp;
        temp = copy;
    }
    return prev;
}
 
function unfold(head)
{
    // for segregating the original linked list into two
    // linked list on the basis of even and odd
    let i = 0;
 
    // For storing the previous node
    // and head node for each linked list
    let prev1 = null, prev2 = null, h1 = head,h2 = head;
 
    while (head != null) {
        if (i % 2 == 0) {
            if (prev1 == null) {
                h1 = head;
                prev1 = head;
            }
            else {
                prev1.next = head;
                prev1 = head;
            }
        }
        else {
            if (prev2 == null) {
                h2 = head;
                prev2 = head;
            }
            else {
                prev2.next = head;
                prev2 = head;
            }
        }
        i++;
        head = head.next;
    }
 
    prev2.next = null;
    let rev = reverse(h2); // reverse the second linked list
    prev1.next = rev; // join the first ll with second one
}
 
function printList(node)
{
    let curr = node;
    while (curr != null) {
        document.write(curr.val," ");
        curr = curr.next;
    }
    document.write("</br>");
}
 
// driver code
let n;
let dummy = new ListNode(-1);
let prev = dummy;
n=5;
let i=0;
let arr = [1, 5, 2, 4, 3]; //Elements in the linkedlist
 
while (i < n) {
    prev.next = new ListNode(arr[i]);
    prev = prev.next;
    i++;
}
 
let head = dummy.next;
 
printList(head);
unfold(head);
printList(head);
 
// This code is contributed by shinjanpatra
 
</script>

Output

1 5 2 4 3 
1 2 3 4 5 

Time Complexity: O(N), where N is the total number of nodes in the linked list. 

Auxiliary Space: O(1)


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!