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.
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.
Node* new_node =
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
(*head_ref == NULL)
Node* temp = *head_ref;
(position == 0)
*head_ref = temp->next;
i = 0; temp != NULL && i < position - 1; i++)
temp = temp->next;
(temp == NULL || temp->next == NULL)
Node *next = temp->next->next;
temp->next = next;
printList( Node *node)
(node != NULL)
cout << node->data <<
node = node->next;
Node* head = NULL;
"Created Linked List: "
cout << "
Linked List after Deletion at position 4: ";
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!