Rearrange a linked list in to alternate first and last element

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

Examples:

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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
    }
}
  
/* Drier 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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.

chevron_right


Python

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


Output:

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

Time Complexity : O(n)



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Article Tags :
Practice Tags :


4


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.