Given a linked list, swap elements in the linked list in pairs
For example, in order to solve the problem of needing to exchange pairs of nodes that exist in the linked list and then print it
Input : 1->2->3->4->5->6->NULL Output : 2->1->4->3->6->5->NULL Input : 1->2->3->4->5->NULL Output : 2->1->4->3->5->NULL Input : 1->NULL Output : 1->NULL
There are two ways to achieve a time complexity of O(N ) where N is the size of the linked list we have provided, so now we will explore both methods
< h2>Iteration methodIn this method we will iterate over the linked list elements and Swap them in pairs until they reach NULL.
Example
#include <bits/stdc++.h> using namespace std; class Node { // node of our list public: int data; Node* next; }; void swapPairwise(Node* head){ Node* temp = head; while (temp != NULL && temp->next != NULL) { // for pairwise swap we need to have 2 nodes hence we are checking swap(temp->data, temp->next->data); // swapping the data temp = temp->next->next; // going to the next pair } } void push(Node** head_ref, int new_data){ // function to push our data in list Node* new_node = new Node(); // creating new node new_node->data = new_data; new_node->next = (*head_ref); // head is pushed inwards (*head_ref) = new_node; // our new node becomes our head } void printList(Node* node){ // utility function to print the given linked list while (node != NULL) { cout << node->data << " "; node = node->next; } } int main(){ Node* head = NULL; push(&head, 5); push(&head, 4); push(&head, 3); push(&head, 2); push(&head, 1); cout << "Linked list before\n"; printList(head); swapPairwise(head); cout << "\nLinked list after\n"; printList(head); return 0; }
Output
Linked list before 1 2 3 4 5 Linked list after 2 1 4 3 5
We will use the same formula in the following method but we will iterate through recursion.
Recursive method
In this method we implement the same logic through recursion.
Example
#include <bits/stdc++.h> using namespace std; class Node { // node of our list public: int data; Node* next; }; void swapPairwise(struct Node* head){ if (head != NULL && head->next != NULL) { // same condition as our iterative swap(head->data, head->next->data); // swapping data swapPairwise(head->next->next); // moving to the next pair } return; // else return } void push(Node** head_ref, int new_data){ // function to push our data in list Node* new_node = new Node(); // creating new node new_node->data = new_data; new_node->next = (*head_ref); // head is pushed inwards (*head_ref) = new_node; // our new node becomes our head } void printList(Node* node){ // utility function to print the given linked list while (node != NULL) { cout << node->data << " "; node = node->next; } } int main(){ Node* head = NULL; push(&head, 5); push(&head, 4); push(&head, 3); push(&head, 2); push(&head, 1); cout << "Linked list before\n"; printList(head); swapPairwise(head); cout << "\nLinked list after\n"; printList(head); return 0; }
Output
Linked list before 1 2 3 4 5 Linked list after 2 1 4 3 5
Explanation of the above code
In this method, we traverse the linked list in pairs. Now when we reach a pair we swap their data and move to the next pair and this is how our program proceeds in both methods.
Conclusion
In this tutorial, we solved pairwise swapping of elements of a given linked list using recursion and iteration. We also learned the C program for this problem and the complete method (general) to solve it. We can write the same program in other languages such as C, java, python and other languages. We hope you found this tutorial helpful.
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