Find the fractional (or n/k - th) node in linked list

Given a singly linked list and a number k, write a function to find the (n/k)-th element, where n is the number of elements in the list. We need to consider ceil value in case of decimals.

Examples:

Input : list = 1->2->3->4->5->6 
        k = 2
Output : 3
Since n = 6 and k = 2, we print (6/2)-th node 
which is 3.

Input : list = 2->7->9->3->5
        k = 3
Output : 7 
Since n is 5 and k is 3, we print ceil(5/3)-th 
node which is 2nd node, i.e., 7.

Recommended: Please solve it on PRACTICE first, before moving on to the solution.

Take two pointers temp and fractionalNode and initialize them with null and head respectively.
For every k jumps of the temp pointer, make one jump of the fractionalNode pointer.

C++

// C++ program to find fractional node in a linked list 
#include <bits/stdc++.h> 
  
/* Linked list node */
struct Node { 
    int data; 
    Node* next; 
}; 
  
/* Function to create a new node with given data */
Node* newNode(int data) 
{ 
    Node* new_node = new Node; 
    new_node->data = data; 
    new_node->next = NULL; 
    return new_node; 
} 
  
/* Function to find fractional node in the linked list */
Node* fractionalNodes(Node* head, int k) 
{ 
    // Corner cases 
    if (k <= 0 || head == NULL) 
        return NULL; 
  
    Node* fractionalNode = NULL; 
      
    // Traverse the given list  
    int i = 0; 
    for (Node* temp = head; temp != NULL; temp = temp->next) { 
  
        // For every k nodes, we move fractionalNode one 
        // step ahead.  
        if (i % k == 0) { 
  
            // First time we see a multiple of k 
            if (fractionalNode == NULL) 
                fractionalNode = head; 
  
            else
                fractionalNode = fractionalNode->next; 
        } 
        i++; 
    } 
    return fractionalNode; 
} 
  
// A utility function to print a linked list 
void printList(Node* node) 
{ 
    while (node != NULL) { 
        printf("%d ", node->data); 
        node = node->next; 
    } 
    printf("\n"); 
} 
  
/* Driver program to test above function */
int main(void) 
{ 
    Node* head = newNode(1); 
    head->next = newNode(2); 
    head->next->next = newNode(3); 
    head->next->next->next = newNode(4); 
    head->next->next->next->next = newNode(5); 
    int k = 2; 
  
    printf("List is "); 
    printList(head); 
  
    Node* answer = fractionalNodes(head, k); 
    printf("\nFractional node is "); 
    printf("%d\n", answer->data); 
  
    return 0; 
} 

Java

// Java program to find fractional node in 
// a linked list 
public class FractionalNodell 
{ 
    /* Linked list node */
    static class Node{ 
        int data; 
        Node next; 
  
        //Constructor 
        Node (int data){ 
            this.data = data; 
        } 
    } 
  
    /* Function to find fractional node in the 
       linked list */
    static Node fractionalNodes(Node head, int k) 
    { 
        // Corner cases 
        if (k <= 0 || head == null) 
            return null; 
  
        Node fractionalNode = null; 
  
        // Traverse the given list 
        int i = 0; 
        for (Node temp = head; temp != null; 
                          temp = temp.next){ 
  
            // For every k nodes, we move 
            // fractionalNode one step ahead. 
            if (i % k == 0){ 
  
                // First time we see a multiple of k 
                if (fractionalNode == null) 
                    fractionalNode = head; 
                else
                    fractionalNode = fractionalNode.next; 
            } 
            i++; 
        } 
        return fractionalNode; 
    } 
  
    // A utility function to print a linked list 
    static void printList(Node node) 
    { 
        while (node != null) 
        { 
            System.out.print(node.data+" "); 
            node = node.next; 
        } 
        System.out.println(); 
    } 
  
    /* Driver program to test above function */
    public static void main(String[] args) { 
        Node head = new Node(1); 
        head.next = new Node(2); 
        head.next.next = new Node(3); 
        head.next.next.next = new Node(4); 
        head.next.next.next.next = new Node(5); 
        int k =2; 
  
        System.out.print("List is "); 
        printList(head); 
  
        Node answer = fractionalNodes(head, k); 
        System.out.println("Fractional node is "+ 
                                      answer.data); 
    } 
} 
// This code is contributed by Sumit Ghosh 

C#

// C# program to find fractional node in 
// a linked list 
using System; 
  
public class FractionalNodell 
{ 
    /* Linked list node */
    public class Node 
    { 
        public int data; 
        public Node next; 
  
        //Constructor 
        public Node (int data) 
        { 
            this.data = data; 
        } 
    } 
  
    /* Function to find fractional node in the 
    linked list */
    static Node fractionalNodes(Node head, int k) 
    { 
        // Corner cases 
        if (k <= 0 || head == null) 
            return null; 
  
        Node fractionalNode = null; 
  
        // Traverse the given list 
        int i = 0; 
        for (Node temp = head; temp != null; 
                        temp = temp.next) 
        { 
  
            // For every k nodes, we move 
            // fractionalNode one step ahead. 
            if (i % k == 0) 
            { 
  
                // First time we see a multiple of k 
                if (fractionalNode == null) 
                    fractionalNode = head; 
                else
                    fractionalNode = fractionalNode.next; 
            } 
            i++; 
        } 
        return fractionalNode; 
    } 
  
    // A utility function to print a linked list 
    static void printList(Node node) 
    { 
        while (node != null) 
        { 
            Console.Write(node.data+" "); 
            node = node.next; 
        } 
        Console.WriteLine(); 
    } 
  
    /* Driver code */
    public static void Main(String[] args)  
    { 
        Node head = new Node(1); 
        head.next = new Node(2); 
        head.next.next = new Node(3); 
        head.next.next.next = new Node(4); 
        head.next.next.next.next = new Node(5); 
        int k =2; 
  
        Console.Write("List is "); 
        printList(head); 
  
        Node answer = fractionalNodes(head, k); 
        Console.WriteLine("Fractional node is "+ 
                                    answer.data); 
    } 
} 
  
// This code is contributed by Rajput-Ji 

Output:


List is 1 2 3 4 5 
Fractional node is 3

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My Personal Notes

Find the fractional (or n/k – th) node in linked list

Recommended: Please solve it on PRACTICE first, before moving on to the solution.

C++

Java

C#

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Recommended: Please solve it on PRACTICE first, before moving on to the solution.

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