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
/* This program swaps the nodes of linked list rather than swapping the field from the nodes. Imagine a case where a node contains many fields, there will be plenty of unnecessary swap calls. */ #include <stdbool.h> #include <stdio.h> #include <stdlib.h> // A linked list node struct Node { int data; struct Node* next; }; /* Function to pairwise swap elements of a linked list */ void pairWiseSwap( struct Node** head) { // If linked list is empty or there // is only one node in list if (*head == NULL || (*head)->next == NULL) return ; // Initialize previous and current // pointers struct Node* prev = *head; struct Node* curr = (*head)->next; // Change head before proceeding *head = curr; // Traverse the list while ( true ) { struct Node* next = curr->next; // Change next of current as // previous node curr->next = prev; // If next NULL or next is the // last node if (next == NULL || next->next == NULL) { prev->next = next; break ; } // Change next of previous to // next next prev->next = next->next; // Update previous and curr prev = next; curr = prev->next; } } /* 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 of 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->6->7 */ 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
/* This program swaps the nodes of linked list rather than swapping the field from the nodes. Imagine a case where a node contains many fields, there will be plenty of unnecessary swap calls. */ #include <stdbool.h> #include <stdio.h> #include <stdlib.h> // A linked list node struct node { int data; struct node* next; }; /* Function to pairwise swap elements of a linked list. It returns head of the modified list, so return value of this node must be assigned */ struct node* pairWiseSwap( struct node* head) { // Base Case: The list is empty or has // only one node if (head == NULL || head->next == NULL) return head; // Store head of list after two nodes struct node* remaining = head->next->next; // Change head struct node* newhead = head->next; // Change next of second node head->next->next = head; // Recur for remaining list and change // next of head head->next = pairWiseSwap(remaining); // Return new head of modified list return newhead; } /* 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 of 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->6->7 */ 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); // NOTE THIS CHANGE 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!
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