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Program to unfold a folded linked list

  • Difficulty Level : Hard
  • Last Updated : 07 Jun, 2021

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;
}
 
// Funcion 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();
    }
 
    // Funcion 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 positio
    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();
    }
 
    // Funcion 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>");
    }
 
// Funcion 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

 

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