Producto de los Nodes de una lista enlazada simple

Dada una lista enlazada simple. La tarea es encontrar el producto de todos los Nodes de la lista enlazada dada.
Ejemplos
 

Input : List = 7->6->8->4->1
Output : Product = 1344
Product of nodes: 7 * 6 * 8 * 4 * 1 = 1344

Input : List = 1->7->3->9->11->5
Output : Product = 10395

Algoritmo
 

  1. Inicialice un puntero ptr con el encabezado de la lista enlazada y una variable de producto con 1.
  2. Comience a recorrer la lista vinculada utilizando un bucle hasta que se atraviesen todos los Nodes.
  3. Multiplique el valor del Node actual por el producto, es decir , producto *= ptr -> datos .
  4. Incremente el puntero al siguiente Node de la lista enlazada, es decir , ptr = ptr ->next .
  5. Repita los dos pasos anteriores hasta llegar al final de la lista enlazada.
  6. Finalmente, devuelva el producto.

A continuación se muestra la implementación del algoritmo anterior: 
 

C++

// C++ implementation to find the product of
// nodes of the Linked List
 
#include <iostream>
using namespace std;
 
// A Linked list node
struct Node {
    int data;
    struct Node* next;
};
 
// Function to insert a node at the
// beginning of the linked list
void push(struct Node** head_ref, int new_data)
{
    /* allocate node */
    struct Node* new_node = new Node;
 
    /* put in the data */
    new_node->data = new_data;
 
    /* link the old list to the new node */
    new_node->next = (*head_ref);
 
    /* move the head to point to the new node */
    (*head_ref) = new_node;
}
 
// Function to find the product of
// nodes of the given linked list
int productOfNodes(struct Node* head)
{
    // Pointer to traverse the list
    struct Node* ptr = head;
 
    int product = 1; // Variable to store product
 
    // Traverse the list and
    // calculate the product
    while (ptr != NULL) {
 
        product *= ptr->data;
        ptr = ptr->next;
    }
 
    // Return the product
    return product;
}
 
// Driver Code
int main()
{
    struct Node* head = NULL;
 
    // create linked list 7->6->8->4->1
    push(&head, 7);
    push(&head, 6);
    push(&head, 8);
    push(&head, 4);
    push(&head, 1);
 
    cout << "Product = " << productOfNodes(head);
 
    return 0;
}

Java

// Java implementation to find the product of
// nodes of the Linked List
class GFG
{
 
// A Linked list node
static class Node
{
    int data;
    Node next;
};
 
// Function to insert a node at the
// beginning of the linked list
static Node push( Node head_ref, int new_data)
{
    // allocate node /
    Node new_node = new Node();
 
    // put in the data /
    new_node.data = new_data;
 
    // link the old list to the new node /
    new_node.next = (head_ref);
 
    // move the head to point to the new node /
    (head_ref) = new_node;
    return head_ref;
}
 
// Function to find the product of
// nodes of the given linked list
static int productOfNodes( Node head)
{
    // Pointer to traverse the list
    Node ptr = head;
 
    int product = 1; // Variable to store product
 
    // Traverse the list and
    // calculate the product
    while (ptr != null)
    {
 
        product *= ptr.data;
        ptr = ptr.next;
    }
 
    // Return the product
    return product;
}
 
// Driver Code
public static void main(String args[])
{
    Node head = null;
 
    // create linked list 7.6.8.4.1
    head = push(head, 7);
    head = push(head, 6);
    head = push(head, 8);
    head = push(head, 4);
    head = push(head, 1);
 
    System.out.println("Product = " + productOfNodes(head));
}
}
 
// This code is contributed by Arnab Kundu

Python3

# Python3 implementation to find the product of
# nodes of the Linked List
import math
 
# A linked List node
class Node:
    def __init__(self,data):
        self.data = data
        self.next = None
 
# Function to insert a node at the
# beginning of the linked list
def push(head,data):
    if not head:
 
        # Return new node
        return Node(data)
     
    # allocate node
    new_node = Node(data)
     
    # link the old list to the new node
    new_node.next = head
     
    # move the head to point to the new node
    head = new_node
    return head
 
# Function to find the product of
# nodes of the given linked list
def productOfNodes(head):
     
    # Pointer to traverse the list
    ptr = head
    product = 1 # Variable to store product
 
    # Traverse the list and
    # calculate the product
    while(ptr):
        product *= ptr.data
        ptr = ptr.next
         
    # Return the product    
    return product
 
# Driver Code
if __name__=='__main__':
    head = None
 
    # create linked list 7->6->8->4->1
    head = push(head,7)
    head = push(head,6)
    head = push(head,8)
    head = push(head,4)
    head = push(head,1)
    print("Product = {}".format(productOfNodes(head)))
 
# This Code is Contributed By Vikash Kumar 37

C#

// C# implementation to find the product of
// nodes of the Linked List
using System;
 
class GFG
{
 
// A Linked list node
public class Node
{
    public int data;
    public Node next;
};
 
// Function to insert a node at the
// beginning of the linked list
static Node push(Node head_ref, int new_data)
{
    // allocate node /
    Node new_node = new Node();
 
    // put in the data /
    new_node.data = new_data;
 
    // link the old list to the new node /
    new_node.next = (head_ref);
 
    // move the head to point to the new node /
    (head_ref) = new_node;
    return head_ref;
}
 
// Function to find the product of
// nodes of the given linked list
static int productOfNodes( Node head)
{
    // Pointer to traverse the list
    Node ptr = head;
 
    int product = 1; // Variable to store product
 
    // Traverse the list and
    // calculate the product
    while (ptr != null)
    {
        product *= ptr.data;
        ptr = ptr.next;
    }
 
    // Return the product
    return product;
}
 
// Driver Code
public static void Main(String []args)
{
    Node head = null;
 
    // create linked list 7.6.8.4.1
    head = push(head, 7);
    head = push(head, 6);
    head = push(head, 8);
    head = push(head, 4);
    head = push(head, 1);
 
    Console.WriteLine("Product = " +
                       productOfNodes(head));
}
}
 
// This code is contributed by Rajput-Ji

Javascript

<script>
// javascript implementation to find the product of
// nodes of the Linked List    
// A Linked list node
class Node {
    constructor(val) {
        this.data = val;
        this.next = null;
    }
}
 
    // Function to insert a node at the
    // beginning of the linked list
    function push(head_ref , new_data) {
        // allocate node /
var new_node = new Node();
 
        // put in the data /
        new_node.data = new_data;
 
        // link the old list to the new node /
        new_node.next = (head_ref);
 
        // move the head to point to the new node /
        (head_ref) = new_node;
        return head_ref;
    }
 
    // Function to find the product of
    // nodes of the given linked list
    function productOfNodes(head) {
        // Pointer to traverse the list
var ptr = head;
 
        var product = 1; // Variable to store product
 
        // Traverse the list and
        // calculate the product
        while (ptr != null) {
 
            product *= ptr.data;
            ptr = ptr.next;
        }
 
        // Return the product
        return product;
    }
 
    // Driver Code
     
var head = null;
 
        // create linked list 7.6.8.4.1
        head = push(head, 7);
        head = push(head, 6);
        head = push(head, 8);
        head = push(head, 4);
        head = push(head, 1);
 
        document.write("Product = " + productOfNodes(head));
 
// This code contributed by aashish1995
</script>
Producción: 

Product = 1344

 

Complejidad de tiempo : O(N), donde N es el número de Nodes en la lista enlazada.
Espacio Auxiliar: O(1) 
 

Publicación traducida automáticamente

Artículo escrito por VishalBachchas y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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