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C Program For Pairwise Swapping Elements Of A Given Linked List

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  • Last Updated : 15 Dec, 2021
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Given a singly linked list, write a function to swap elements pairwise.

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

For example, if the linked list is 1->2->3->4->5 then the function should change it to 2->1->4->3->5, and if the linked list is then the function should change it to.

METHOD 1 (Iterative): 
Start from the head node and traverse the list. While traversing swap data of each node with its next node’s data. 
Below is the implementation of the above approach:

C




/* C program to pairwise swap elements 
   in a given linked list */
#include <stdio.h>
#include <stdlib.h>
  
// A linked list node 
struct Node 
{
    int data;
    struct Node* next;
};
  
/* Function to swap two integers 
   at addresses a and b */
void swap(int* a, int* b);
  
/* Function to pairwise swap elements 
   of a linked list */
void pairWiseSwap(struct Node* head)
{
    struct Node* temp = head;
  
    /* Traverse further only if there 
       are at-least two nodes left */
    while (temp != NULL && 
           temp->next != NULL) 
    {
        /* Swap data of node with its 
           next node's data */
        swap(&temp->data, 
             &temp->next->data);
  
        // Move temp by 2 for the 
        // next pair 
        temp = temp->next->next;
    }
}
  
// UTILITY FUNCTIONS 
// Function to swap two integers 
void swap(int* a, int* b)
{
    int temp;
    temp = *a;
    *a = *b;
    *b = temp;
}
  
/* Function to add a node at the 
   beginning of Linked List */
void push(struct Node** head_ref, int 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 point to the 
    // new node 
    (*head_ref) = new_node;
}
  
/* Function to print nodes in a 
   given linked list */
void printList(struct Node* node)
{
    while (node != NULL) 
    {
        printf("%d ", node->data);
        node = node->next;
    }
}
  
// Driver code
int main()
{
    struct Node* start = NULL;
  
    /* The constructed linked list is: 
       1->2->3->4->5 */
    push(&start, 5);
    push(&start, 4);
    push(&start, 3);
    push(&start, 2);
    push(&start, 1);
  
    printf(
    "Linked list before calling pairWiseSwap()");
    printList(start);
  
    pairWiseSwap(start);
  
    printf(
    "Linked list after calling pairWiseSwap()");
    printList(start);
  
    return 0;
}


Output:

Linked list before calling pairWiseSwap()
1 2 3 4 5 
Linked list after calling pairWiseSwap()
2 1 4 3 5 

Time complexity: O(n)

METHOD 2 (Recursive): 
If there are 2 or more than 2 nodes in Linked List then swap the first two nodes and recursively call for rest of the list.
Below image is a dry run of the above approach:

Below is the implementation of the above approach:

C




/* Recursive function to pairwise swap 
   elements of a linked list */
void pairWiseSwap(struct node* head)
{
    /* There must be at-least two nodes 
       in the list */
    if (head != NULL && head->next != NULL) 
    {
        /* Swap the node's data with data 
           of next node */
        swap(head->data, head->next->data);
  
        /* Call pairWiseSwap() for rest of 
           the list */
        pairWiseSwap(head->next->next);
    }
}


Time complexity: O(n)
The solution provided there swaps data of nodes. If data contains many fields, there will be many swap operations. See this for an implementation that changes links rather than swapping data.

Please refer complete article on Pairwise swap elements of a given linked list for more details!


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