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++
// 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 functionchanges 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 listvoid 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;} |
Java
// Java program to rearrange a linked list in given mannerimport 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 listdef 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 1.2.3.4head = 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 |
Output:
Given linked list 1 2 3 4 5 After rearrangement 1 5 2 4 3
Time Complexity : O(n)
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