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++
#include <iostream>
using namespace std;
class Node
{
public :
int data;
Node *next;
};
void push(Node** head_ref, int new_data)
{
Node* new_node = new Node();
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
void deleteNode(Node **head_ref, int position)
{
if (*head_ref == NULL)
return ;
Node* temp = *head_ref;
if (position == 0)
{
*head_ref = temp->next;
free (temp);
return ;
}
for ( int i = 0; temp != NULL && i < position - 1; i++)
temp = temp->next;
if (temp == NULL || temp->next == NULL)
return ;
Node *next = temp->next->next;
free (temp->next);
temp->next = next;
}
void printList( Node *node)
{
while (node != NULL)
{
cout << node->data << " " ;
node = node->next;
}
}
int main()
{
Node* head = NULL;
push(&head, 7);
push(&head, 1);
push(&head, 3);
push(&head, 2);
push(&head, 8);
cout << "Created Linked List: " ;
printList(head);
deleteNode(&head, 4);
cout << "
Linked List after Deletion at position 4: ";
printList(head);
return 0;
}
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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!