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.


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.

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





// 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) {
        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();
        else {
            temp->data = d.back(); 
        temp = temp->next; // increase temp
/* 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";
    cout << "\nAfter rearrangement\t";
    return 0;



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

Time Complexity : O(n)

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