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C++ Program For Deleting A Linked List Node At A Given Position

Last Updated : 16 Jun, 2022
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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 <iostream>
using namespace std;
 
// A linked list node
class Node
{
    public:
    int data;
    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(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;
}
 
// Given a reference (pointer to pointer) to
// the head of a list and a position, deletes
// the node at the given position
void deleteNode(Node **head_ref, int position)
{
     
    // If linked list is empty
    if (*head_ref == NULL)
        return;
     
    // Store head node
    Node* temp = *head_ref;
 
    // If head needs to be removed
    if (position == 0)
    {
         
        // Change head
        *head_ref = temp->next;
         
        // Free old head
        free(temp);            
        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
     Node *next = temp->next->next;
 
    // Unlink the node from linked list
    free(temp->next); // Free memory
     
    // Unlink the deleted node from list
    temp->next = next;
}
 
// This function prints contents of linked
// list starting from the given node
void printList( Node *node)
{
    while (node != NULL)
    {
        cout << node->data << " ";
        node = node->next;
    }
}
 
// Driver code
int main()
{
     
    // Start with the empty list
    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;
}
 
// This code is contributed by premsai2030


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!



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