Given a Singly Linked List, starting from the second node delete all alternate nodes of it. For example, if the given linked list is 1->2->3->4->5 then your function should convert it to 1->3->5, and if the given linked list is 1->2->3->4 then convert it to 1->3.
Method 1 (Iterative):
Keep track of previous of the node to be deleted. First, change the next link of the previous node and iteratively move to the next node.
// C program to remove alternate nodes // of a linked list #include<stdio.h> #include<stdlib.h> // A linked list node struct Node
{ int data;
struct Node *next;
}; // Deletes alternate nodes of a list // starting with head void deleteAlt( struct Node *head)
{ if (head == NULL)
return ;
// Initialize prev and node to
// be deleted
struct Node *prev = head;
struct Node *node = head->next;
while (prev != NULL &&
node != NULL)
{
// Change next link of previous
// node
prev->next = node->next;
// Free memory
free (node);
// Update prev and node
prev = prev->next;
if (prev != NULL)
node = prev->next;
}
} // UTILITY FUNCTIONS TO TEST // fun1() and fun2() /* Given a reference (pointer to pointer) to the head of a list and an int, push
a new node on the front of the 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()
{ // Start with the empty list
struct Node* head = NULL;
/* Using push() to construct list
1->2->3->4->5 */
push(&head, 5);
push(&head, 4);
push(&head, 3);
push(&head, 2);
push(&head, 1);
printf ( "\nList before calling deleteAlt() \n" );
printList(head);
deleteAlt(head);
printf ( "\nList after calling deleteAlt() \n" );
printList(head);
return 0;
} |
Output:
List before calling deleteAlt() 1 2 3 4 5 List after calling deleteAlt() 1 3 5
Time Complexity: O(n) where n is the number of nodes in the given Linked List.
Auxiliary Space: O(1) because it is using constant space
Method 2 (Recursive):
Recursive code uses the same approach as method 1. The recursive code is simple and short but causes O(n) recursive function calls for a linked list of size n.
// Deletes alternate nodes of a list // starting with head void deleteAlt( struct Node *head)
{ if (head == NULL)
return ;
struct Node *node = head->next;
if (node == NULL)
return ;
// Change the next link of head
head->next = node->next;
// Free memory allocated for node
free (node);
// Recursively call for the new
// next of head
deleteAlt(head->next);
} |
Time Complexity: O(n)
Auxiliary space: O(n) for call stack because using recursion
Please refer complete article on Delete alternate nodes of a Linked List for more details!