Remove last node of the linked list

Given a linked list, the task is to remove the last node of the linked list and update the head pointer of the linked list.

Examples:

Input : 1 -> 2 -> 3 -> 4 -> 5 -> NULL
Output : 1 -> 2 -> 3 -> 4 -> NULL

Input : 2 -> 4 -> 6 -> 8 -> 33 -> 67 -> NULL
Output : 2 -> 4 -> 6 -> 8 -> 33 -> NULL


To remove last node, we need to make the next of second last node as NULL and delete memory allocated for last node.

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// CPP program to remove last node of
// linked list.
#include <iostream>
using namespace std;
  
/* Link list node */
struct Node {
    int data;
    struct Node* next;
};
  
/* Function to remove the last node  
   of the linked list */
Node* removeLastNode(struct Node* head)
{
    if (head == NULL)
        return NULL;
  
    if (head->next == NULL)
    {
        delete head;
        return NULL;
    }
  
    // Find the second last node
    Node* second_last = head;
    while (second_last->next->next != NULL)
        second_last = second_last->next;
  
    // Delete last node
    delete (second_last->next);
  
    // Change next of second last
    second_last->next = NULL;
  
    return head;
}
  
// Function to push node at head
void push(struct Node** head_ref, int new_data)
{
    struct Node* new_node = new Node;
    new_node->data = new_data;
    new_node->next = (*head_ref);
    (*head_ref) = new_node;
}
  
// Driver code
int main()
{
    /* Start with the empty list */
    Node* head = NULL;
  
    /* Use push() function to construct   
       the below list 8 -> 23 -> 11 -> 29 -> 12 */
    push(&head, 12);
    push(&head, 29);
    push(&head, 11);
    push(&head, 23);
    push(&head, 8);
  
    head = removeLastNode(head);
    for (Node* temp = head; temp != NULL; temp = temp->next)
        cout << temp->data << " ";
  
    return 0;
}

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Output:

8 23 11 29

Time Complexity : O(n)



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Improved By : VishalBachchas