查找给定链接列表的前k个节点的乘积

给定一个指向单链表头的指针和一个整数k 。任务是找到链表中前k个节点的乘积。

例子：

Input: 10 -> 6 -> 8 -> 4 -> 12, k = 2
Output: 60
10 * 6 = 60

Input: 15 -> 7 -> 9 -> 5 -> 16 -> 14, k = 4
Output: 4725
15 * 7 * 9 * 5 = 4725

方法：设置prod = 1 (必需产品)，计数= 0 (遍历的节点数)。现在，开始从左到右遍历链表的节点，并在count 时，对每个遍历的节点更新count = count +1和prod = prod * currNode->数据。最后输出prod的值。

下面是上述方法的实现：

C++

// C++ program to find the product of first
// 'k' nodes of the Linked List
#include 
using namespace std;
#define ll long long int
  
/* 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 return the product of
// first k nodes of the given linked list
ll product(struct Node* head, int k)
{
    if (k <= 0)
        return 0;
  
    ll prod = 1;
    int i = 0;
    Node* node = head;
  
    // Traverse the list from left to right
    while (i < k) {
  
        // Update product
        prod = prod * node->data;
  
        // Move to the next node
        node = node->next;
        i++;
    }
  
    // Return the required product
    return prod;
}
  
// Driver code
int main()
{
    struct Node* head = NULL;
  
    // Linked list 10 -> 6 -> 8 -> 4 -> 12
    push(&head, 12);
    push(&head, 4);
    push(&head, 8);
    push(&head, 6);
    push(&head, 10);
  
    int k = 2;
    cout << product(head, k);
    return 0;
}

Java

// Java program to find the product of first
// 'k' nodes of the Linked List
class Solution
{
  
/* 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 return the product of
// first k nodes of the given linked list
static long product( Node head, int k)
{
    if (k <= 0)
        return 0;
  
    long prod = 1;
    int i = 0;
    Node node = head;
  
    // Traverse the list from left to right
    while (i < k)
    {
  
        // Update product
        prod = prod * node.data;
  
        // Move to the next node
        node = node.next;
        i++;
    }
  
    // Return the required product
    return prod;
}
  
// Driver code
public static void main(String args[])
{
    Node head = new Node();
  
    // Linked list 10 . 6 . 8 . 4 . 12
    head=push(head, 12);
    head=push(head, 4);
    head=push(head, 8);
    head=push(head, 6);
    head=push(head, 10);
  
    int k = 2;
    System.out.println( product(head, k));
      
}
}
  
// This code is contributed by Arnab Kundu

Python3

# Python3 program to find the product 
# of first 'k' nodes of the Linked List
import math
  
#define ll long long 
  
# 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_ref, new_data):
      
    # allocate node 
    new_node = Node(new_data)
  
    # 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 po to the new node 
    head_ref = new_node
    return head_ref
  
# Function to return the product of
# first k nodes of the given linked list
def product(head, k):
    if (k <= 0):
        return 0
  
    prod = 1
    i = 0
    node = head
  
    # Traverse the list from left to right
    while (i < k):
  
        # Update product
        prod = prod * node.data
  
        # Move to the next node
        node = node.next
        i = i + 1
      
    # Return the required product
    return prod
  
# Driver code
if __name__=='__main__':
    head = None
  
    # Linked list 10 . 6 . 8 . 4 . 12
    head = push(head, 12);
    head = push(head, 4);
    head = push(head, 8);
    head = push(head, 6);
    head = push(head, 10);
  
    k = 2
    print(product(head, k))
      
# This code is contributed by Srathore

C#

// C# program to find the product of first
// 'k' nodes of the Linked List
using System;
  
class GFG
{
  
/* A Linked list node */
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 return the product of
// first k nodes of the given linked list
static long product( Node head, int k)
{
    if (k <= 0)
        return 0;
  
    long prod = 1;
    int i = 0;
    Node node = head;
  
    // Traverse the list from left to right
    while (i < k)
    {
  
        // Update product
        prod = prod * node.data;
  
        // Move to the next node
        node = node.next;
        i++;
    }
  
    // Return the required product
    return prod;
}
  
// Driver code
public static void Main()
{
    Node head = new Node();
  
    // Linked list 10 . 6 . 8 . 4 . 12
    head=push(head, 12);
    head=push(head, 4);
    head=push(head, 8);
    head=push(head, 6);
    head=push(head, 10);
  
    int k = 2;
    Console.WriteLine( product(head, k));
}
}
  
// This code is contributed by PrinciRaj1992

输出：

60

想要从精选的最佳视频中学习并解决问题，请查看有关从基础到高级C++的C++基础课程以及有关语言和STL的C++ STL课程。要完成从学习语言到DS Algo等的更多准备工作，请参阅“完整面试准备课程” 。