C Exercises: Nodes from the end of a singly linked list

Last update on August 02 2025 12:55:35 (UTC/GMT +8 hours)

20. Nth Node from End Variants

Write a C program to get the n number of nodes from the end of a singly linked list.

Sample Solution:

C Code:

#include<stdio.h>
#include <stdlib.h>

// Define a structure for a Node in a singly linked list
struct Node {
    int data;
    struct Node* next;
};

// Function to create a new node with given data
struct Node* new_Node(int data) {
    // Allocate memory for a new node
    struct Node* node = (struct Node*) malloc(sizeof(struct Node));
    // Set the data for the new node
    node->data = data;
    // Set the next pointer of the new node to NULL
    node->next = NULL;
    return node;
}

// Function to display the elements of a linked list
void displayList(struct Node* head) {
    while (head) {
        printf("%d ", head->data);
        head = head->next;
    }
    printf("\n");
}

// Function to find the Nth node from the end of a linked list
struct Node* Nodes_From_End(struct Node* head, int n) {
    struct Node *slow_ptr = head, *fast_ptr = head;
    int ctr = 0;

    // Move the fast pointer n positions ahead
    while (fast_ptr != NULL) {
        fast_ptr = fast_ptr->next;
        // Once fast pointer is n nodes ahead, move the slow pointer
        if (ctr >= n) {
            slow_ptr = slow_ptr->next;
        }
        ctr++;
    }

    // If the number of nodes is less than N, return NULL
    if (ctr < n)
        return NULL;
    return slow_ptr; // Return the node at Nth position from the end
}

// Main function where the execution starts
int main() {
    struct Node* result;
    struct Node* list1 = new_Node(1);
    list1->next = new_Node(3);
    list1->next->next = new_Node(5);
    list1->next->next->next = new_Node(7);  
    list1->next->next->next->next = new_Node(9); 
    list1->next->next->next->next->next = new_Node(11); 

    printf("Original list:\n");
    displayList(list1); // Display the original list

    // Finding the last N nodes from the end of the list for different values of N
    printf("Last 2 nodes from the end of the said singly list:\n");
    result = Nodes_From_End(list1, 2);
    displayList(result);

    printf("Last 3 nodes from the end of the said singly list:\n");
    result = Nodes_From_End(list1, 3);
    displayList(result);

    printf("Last 4 nodes from the end of the said singly list:\n");
    result = Nodes_From_End(list1, 4);
    displayList(result);

    printf("Last 1 node from the end of the said singly list:\n");
    result = Nodes_From_End(list1, 1);
    displayList(result);

    printf("Last 5 node from the end of the said singly list:\n");
    result = Nodes_From_End(list1, 5);
    displayList(result);

    return 0; // Indicates successful completion of the program
}

Sample Output:

Original list:
1 3 5 11 
Last 2 nodes from the end of the said singly list:
5 11 
Last 3 nodes from the end of the said singly list:
3 5 11 
Last 4 nodes from the end of the said singly list:
1 3 5 11 
Last 1 node from the end of the said singly list:
11 
Last 5 node from the end of the said singly list:

Flowchart :

For more Practice: Solve these Related Problems:

Write a C program to find the nth node from the end of a singly linked list using the two-pointer approach.
Write a C program to extract the last n nodes from a linked list and store them in a new linked list.
Write a C program to recursively determine the nth node from the end of a singly linked list.
Write a C program to extract nodes from the end of a linked list that satisfy a specific condition (e.g., being a prime number).

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