C Program For Deleting A Linked List Node At A Given Position
Given a singly linked list and a position, delete a linked list node at the given position.
Example:
Input: position = 1, Linked List = 8->2->3->1->7 Output: Linked List = 8->3->1->7 Input: position = 0, Linked List = 8->2->3->1->7 Output: Linked List = 2->3->1->7
If the node to be deleted is the root, simply delete it. To delete a middle node, we must have a pointer to the node previous to the node to be deleted. So if positions are not zero, we run a loop position-1 times and get a pointer to the previous node.
Below is the implementation of the above idea.
C
// A complete working C program to delete a node in a linked list // at a given position #include <stdio.h> #include <stdlib.h> // A linked list node struct Node { int data; struct Node *next; }; /* Given a reference (pointer to pointer) to the head of a list and an int inserts a new node on the front of the list. */ 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; } /* Given a reference (pointer to pointer) to the head of a list and a position, deletes the node at the given position */ void deleteNode( struct Node **head_ref, int position) { // If linked list is empty if (*head_ref == NULL) return ; // Store head node struct Node* temp = *head_ref; // If head needs to be removed if (position == 0) { *head_ref = temp->next; // Change head free (temp); // free old head return ; } // Find previous node of the node to be deleted for ( int i=0; temp!=NULL && i<position-1; i++) temp = temp->next; // If position is more than number of nodes if (temp == NULL || temp->next == NULL) return ; // Node temp->next is the node to be deleted // Store pointer to the next of node to be deleted struct Node *next = temp->next->next; // Unlink the node from linked list free (temp->next); // Free memory temp->next = next; // Unlink the deleted node from list } // This function prints contents of linked list starting from // the given node void printList( struct Node *node) { while (node != NULL) { printf ( " %d " , node->data); node = node->next; } } /* Driver program to test above functions*/ int main() { /* Start with the empty list */ struct Node* head = NULL; push(&head, 7); push(&head, 1); push(&head, 3); push(&head, 2); push(&head, 8); puts ( "Created Linked List: " ); printList(head); deleteNode(&head, 4); puts (" Linked List after Deletion at position 4: "); printList(head); return 0; } |
Output:
Created Linked List: 8 2 3 1 7 Linked List after Deletion at position 4: 8 2 3 1
Time Complexity: O(n), where n represents the length of the given linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Please refer complete article on Delete a Linked List node at a given position for more details!