C Program For Pairwise Swapping Elements Of A Given Linked List By Changing Links
Given a singly linked list, write a function to swap elements pairwise. For example, if the linked list is 1->2->3->4->5->6->7 then the function should change it to 2->1->4->3->6->5->7, and if the linked list is 1->2->3->4->5->6 then the function should change it to 2->1->4->3->6->5
This problem has been discussed here. The solution provided there swaps data of nodes. If data contains many fields, there will be many swap operations. So changing links is a better idea in general. Following is the implementation that changes links instead of swapping data.
C
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
struct Node
{
int data;
struct Node* next;
};
void pairWiseSwap( struct Node** head)
{
if (*head == NULL ||
(*head)->next == NULL)
return ;
struct Node* prev = *head;
struct Node* curr = (*head)->next;
*head = curr;
while ( true )
{
struct Node* next = curr->next;
curr->next = prev;
if (next == NULL ||
next->next == NULL)
{
prev->next = next;
break ;
}
prev->next = next->next;
prev = next;
curr = prev->next;
}
}
void push( struct Node** head_ref,
int new_data)
{
struct Node* new_node =
( struct Node*) malloc ( sizeof ( struct Node));
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
void printList( struct Node* node)
{
while (node != NULL)
{
printf ( "%d " ,
node->data);
node = node->next;
}
}
int main()
{
struct Node* start = NULL;
push(&start, 7);
push(&start, 6);
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);
getchar ();
return 0;
}
|
Output:
Linked list before calling pairWiseSwap() 1 2 3 4 5 6 7
Linked list after calling pairWiseSwap() 2 1 4 3 6 5 7
Time Complexity: The time complexity of the above program is O(n) where n is the number of nodes in a given linked list. The while loop does a traversal of the given linked list.
Auxiliary Space: O(1)
Following is the recursive implementation of the same approach. We change the first two nodes and recur for the remaining list. Thanks to geek and omer salem for suggesting this method.
C
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
struct node
{
int data;
struct node* next;
};
struct node* pairWiseSwap( struct node* head)
{
if (head == NULL ||
head->next == NULL)
return head;
struct node* remaining =
head->next->next;
struct node* newhead = head->next;
head->next->next = head;
head->next = pairWiseSwap(remaining);
return newhead;
}
void push( struct node** head_ref,
int new_data)
{
struct node* new_node =
( struct node*) malloc ( sizeof ( struct node));
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
void printList( struct node* node)
{
while (node != NULL)
{
printf ( "%d " , node->data);
node = node->next;
}
}
int main()
{
struct node* start = NULL;
push(&start, 7);
push(&start, 6);
push(&start, 5);
push(&start, 4);
push(&start, 3);
push(&start, 2);
push(&start, 1);
printf ("
Linked list before calling pairWiseSwap() ");
printList(start);
start = pairWiseSwap(start);
printf ("
Linked list after calling pairWiseSwap() ");
printList(start);
return 0;
}
|
Output:
Linked list before calling pairWiseSwap() 1 2 3 4 5 6 7
Linked list after calling pairWiseSwap() 2 1 4 3 6 5 7
Time Complexity: O(n)
Auxiliary Space: O(n)
Please refer complete article on Pairwise swap elements of a given linked list by changing links for more details!
Last Updated :
10 Nov, 2022
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