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Pairwise Swap Elements of a Given Linked List in C++
To solve a problem in which we are required to swap the pairwise nodes present in a linked list and then print it, for example
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 approach the solution both to have a time complexity of O(N), where N is the size of our provided linked list, so now we are going to explore both of the approaches
Iterative Approach
We will iterate through the linked list elements in this approach, and pairwise swap them 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 our following approach, but we will iterate through recursion.
Recursive Approach
In this approach, we are implementing the same logic with 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 approach, we traverse through our linked list in pairs. Now, as we reach a pair, we swap their data and move to the next pair, and that’s how our program proceeds in both methods.
Conclusion
In this tutorial, we solve Pairwise swap elements of a given linked list using recursion and iteration. We also learned the C++ program for this problem and the complete approach (Normal) by which we solved this problem. We can write the same program in other languages such as C, java, python, and other languages. We hope you find this tutorial helpful.