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Rearrange a linked list in to alternate first and last element

  • Difficulty Level : Hard
  • Last Updated : 30 Sep, 2021

Given a linked list. arrange the linked list in manner of alternate first and last element.

Examples: 

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Input : 1->2->3->4->5->6->7->8
Output :1->8->2->7->3->6->4->5

Input :10->11->15->13
Output :10->13->11->15

We have discussed three different solution in Rearrange a given linked list in-place.
In this post a different Deque based solution is discussed.



Method 
1) Create an empty deque 
2) Insert all element from the linked list to the deque 
3) Insert the element back to the linked list from deque in alternate fashion i.e first then last and so on 

C++




// CPP program to rearrange a linked list in given manner
#include <bits/stdc++.h>
using namespace std;
 
/* Link list node */
struct Node {
    int data;
    struct Node* next;
};
 
/* Function to reverse the linked list */
void arrange(struct Node* head)
{
    struct Node* temp = head;
    deque<int> d; // defining a deque
 
    // push all the elements of linked list in to deque
    while (temp != NULL) {
        d.push_back(temp->data);
        temp = temp->next;
    }
 
    // Alternatively push the first and last elements
    // from deque to back to the linked list and pop
    int i = 0;
    temp = head;
    while (!d.empty()) {
        if (i % 2 == 0) {
            temp->data = d.front();
            d.pop_front();
        }
        else {
            temp->data = d.back();
            d.pop_back();
        }
        i++;
        temp = temp->next; // increase temp
    }
}
 
/*UTILITY FUNCTIONS*/
/* Push a node to linked list. Note that this function
changes the head */
void push(struct Node** head_ref, char new_data)
{
    /* allocate node */
    struct Node* new_node = (struct Node*)malloc(sizeof(struct Node));
 
    /* put in the data */
    new_node->data = new_data;
 
    /* link the old list off the new node */
    new_node->next = (*head_ref);
 
    /* move the head to pochar to the new node */
    (*head_ref) = new_node;
}
 
// printing the linked list
void printList(struct Node* head)
{
    struct Node* temp = head;
    while (temp != NULL) {
        printf("%d  ", temp->data);
        temp = temp->next;
    }
}
 
/* Driver program to test above function*/
int main()
{
    // Let us create linked list 1->2->3->4
    struct Node* head = NULL;
 
    push(&head, 5);
    push(&head, 4);
    push(&head, 3);
    push(&head, 2);
    push(&head, 1);
    cout << "Given linked list\t";
    printList(head);
    arrange(head);
    cout << "\nAfter rearrangement\t";
    printList(head);
    return 0;
}

Java




// Java program to rearrange a linked list in given manner
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG
{
    /* Link list node */
    static class Node
    {
        int data;
        Node next;
        Node(int data)
        {
            this.data = data;
            next = null;
        }
    }
     
    // printing the linked list
    static void printList(Node head)
    {
        Node temp = head;
        while (temp != null)
        {
            System.out.print(temp.data + " ");
            temp = temp.next;
        }
    }
 
    /* Function to reverse the linked list */
    static void arrange(Node head)
    {
        // defining a deque
        Deque<Integer> deque = new ArrayDeque<>();
        Node temp = head;
         
        // push all the elements of linked list in to deque
        while(temp != null)
        {
            deque.addLast(temp.data);
            temp = temp.next;
        }
        temp = head;
        int i = 0;
         
        // Alternatively push the first and last elements
        // from deque to back to the linked list and pop
        while(!deque.isEmpty())
        {
            if(i % 2 == 0)
            {
                temp.data = deque.removeFirst();
            }
            else
            {
                temp.data = deque.removeLast();
            }
            i++;
            temp = temp.next;
        }
    }
 
    // Driver code
    public static void main (String[] args)
    {
        // Let us create linked list 1->2->3->4->5
        Node head = null;
         
        head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);
        head.next.next.next = new Node(4);
        head.next.next.next.next = new Node(5);
 
        System.out.println("Given linked list");
        printList(head);
        arrange(head);
        System.out.println("\nAfter rearrangement");
        printList(head);
    }
}
 
// This code is contributed by nobody_cares.

Python




# Python program to rearrange
# a linked list in given manner
 
# Link list node
class Node:
     
    def __init__(self, data):
        self.data = data
        self.next = None
 
# Function to reverse the linked list
def arrange( head):
 
    temp = head
     
    # defining a deque
    d = []
     
    # push all the elements of linked list in to deque
    while (temp != None) :
        d.append(temp.data)
        temp = temp.next
     
    # Alternatively push the first and last elements
    # from deque to back to the linked list and pop
    i = 0
    temp = head
    while (len(d) > 0) :
        if (i % 2 == 0) :
            temp.data = d[0]
            d.pop(0)
         
        else :
            temp.data = d[-1]
            d.pop()
         
        i = i + 1
 
        # increase temp
        temp = temp.next
         
    return head
     
# UTILITY FUNCTIONS
# Push a node to linked list. Note that this function
# changes the head
def push( head_ref, new_data):
 
    # allocate node
    new_node = Node(0)
 
    # put in the data
    new_node.data = new_data
 
    # link the old list off the new node
    new_node.next = (head_ref)
 
    # move the head to pochar to the new node
    (head_ref) = new_node
    return head_ref
 
# printing the linked list
def printList( head):
 
    temp = head
    while (temp != None) :
        print( temp.data,end=" ")
        temp = temp.next
 
# Driver program to test above function
 
# Let us create linked list 1.2.3.4
head = None
 
head = push(head, 5)
head = push(head, 4)
head = push(head, 3)
head = push(head, 2)
head = push(head, 1)
print("Given linked list\t")
printList(head)
head = arrange(head)
print( "\nAfter rearrangement\t")
printList(head)
 
# This code is contributed by Arnab Kundu

Javascript




<script>
 
// Javascript program to rearrange a linked list in given manner
 
/* Link list node */
class Node {
 
    constructor()
    {
        this.data = 0;
        this.next = null;
    }
};
 
/* Function to reverse the linked list */
function arrange( head)
{
    var temp = head;
    var d = []; // defining a deque
 
    // push all the elements of linked list in to deque
    while (temp != null) {
        d.push(temp.data);
        temp = temp.next;
    }
 
    // Alternatively push the first and last elements
    // from deque to back to the linked list and pop
    var i = 0;
    temp = head;
    while (d.length!=0) {
        if (i % 2 == 0) {
            temp.data = d[0];
            d.shift();
        }
        else {
            temp.data = d[d.length-1];
            d.pop();
        }
        i++;
        temp = temp.next; // increase temp
    }
}
 
/*UTILITY FUNCTIONS*/
/* Push a node to linked list. Note that this function
changes the head */
function push(head_ref, new_data)
{
    /* allocate node */
    var new_node = new Node();
 
    /* put in the data */
    new_node.data = new_data;
 
    /* link the old list off the new node */
    new_node.next = (head_ref);
 
    /* move the head to pochar to the new node */
    (head_ref) = new_node;
 
    return head_ref;
}
 
// printing the linked list
function printList(head)
{
    var temp = head;
    while (temp != null) {
        document.write(temp.data+ " ");
        temp = temp.next;
    }
}
 
/* Driver program to test above function*/
// Let us create linked list 1.2.3.4
var head = null;
head = push(head, 5);
head = push(head, 4);
head = push(head, 3);
head = push(head, 2);
head = push(head, 1);
document.write( "Given linked list  ");
printList(head);
arrange(head);
document.write( "<br>After rearrangement  ");
printList(head);
 
 
</script>

Output:  

Given linked list    1 2 3 4 5 
After rearrangement    1 5 2 4 3

Time Complexity : O(n)
 




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