双链表中所有素数节点的乘积

给定一个包含 N 个节点的双向链表。任务是找到所有主要节点的乘积。
例子：

Input: List = 15 <=> 16 <=> 6 <=> 7 <=> 17 
Output: Product of Prime Nodes: 119
Input: List = 5 <=> 3 <=> 4 <=> 2 <=> 9 
Output: Product of Prime Nodes: 30 
 

方法：

• 用链表的头部初始化一个指针 temp，用 1 初始化一个乘积变量。
• 使用循环开始遍历链表，直到遍历完所有节点。
• 如果节点值为素数，则将当前节点的值乘以乘积，即乘积 *= current_node-> 数据。
• 增加指向链表下一个节点的指针，即 temp = temp -> next。
• 退回产品。

下面是上述方法的实现：

C++

// C++ implementation to product all
// prime nodes from the doubly
// linked list
#include 
 
using namespace std;
 
// Node of the doubly linked list
struct Node {
    int data;
    Node *prev, *next;
};
 
// function to insert a node at the beginning
// of the Doubly Linked List
void push(Node** head_ref, int new_data)
{
    // allocate node
    Node* new_node = (Node*)malloc(sizeof(struct Node));
 
    // put in the data
    new_node->data = new_data;
 
    // since we are adding at the beginning,
    // prev is always NULL
    new_node->prev = NULL;
 
    // link the old list off the new node
    new_node->next = (*head_ref);
 
    // change prev of head node to new node
    if ((*head_ref) != NULL)
        (*head_ref)->prev = new_node;
 
    // move the head to point to the new node
    (*head_ref) = new_node;
}
 
// Function to check if a number is prime
bool isPrime(int n)
{
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
 
    // This is checked so that we can skip
    // middle five numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
        return false;
 
    for (int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
 
    return true;
}
 
// function to product all prime nodes
// from the doubly linked list
int prodOfPrime(Node** head_ref)
{
    Node* ptr = *head_ref;
    Node* next;
    // variable prod = 1 for multiplying nodes value
    int prod = 1;
    // traverse list till last node
    while (ptr != NULL) {
        next = ptr->next;
        // if number is prime then
        // multiply to product
        if (isPrime(ptr->data))
            prod = prod * ptr->data;
        ptr = next;
    }
 
    // return product
    return prod;
}
 
// Driver program
int main()
{
    // start with the empty list
    Node* head = NULL;
 
    // create the doubly linked list
    // 15 <-> 16 <-> 7 <-> 6 <-> 17
    push(&head, 17);
    push(&head, 6);
    push(&head, 7);
    push(&head, 16);
    push(&head, 15);
    int prod = prodOfPrime(&head);
 
    cout << "Product of Prime Nodes : " << prod;
 
    return 0;
}

Java

// Java implementation to product all
// prime nodes from the doubly
// linked list
 
// Node of the doubly linked list
class Node
{
    int data;
    Node next, prev;
 
    Node(int d)
    {
        data = d;
        next = null;
        prev = null;
    }
}
 
class DLL
{
    // function to insert a node at the beginning
    // of the Doubly Linked List
    static Node push(Node head, int data)
    {
        Node newNode = new Node(data);
        newNode.next = head;
        newNode.prev = null;
        if (head != null)
            head.prev = newNode;
        head = newNode;
 
        return head;
    }
 
    // Function to check if a number is prime
    static boolean isPrime(int n)
    {
        // Corner cases
        if (n <= 1)
            return false;
        if (n <= 3)
            return true;
 
        // This is checked so that we can skip
        // middle five numbers in below loop
        if (n % 2 == 0 || n % 3 == 0)
            return false;
 
        for (int i = 5; i * i <= n; i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;
 
        return true;
    }
 
    // function to product all prime nodes
    // from the doubly linked list
    static int prodOfPrime(Node node)
    {
        // variable prod = 1 for multiplying nodes value
        int prod = 1;
         
        // traverse list till last node
        while (node != null)
        {
            // check is node value is Prime
            // if true then multiply to prod
            if (isPrime(node.data))
                prod *= node.data;
            node = node.next;
        }
        // return product
        return prod;
    }
 
    // Driver Program
    public static void main(String[] args)
    {
        // Start with empty list
        Node head = null;
 
        // create the doubly linked list
        // 15 <-> 16 <-> 7 <-> 6 <-> 17
        head = push(head, 17);
        head = push(head, 7);
        head = push(head, 6);
        head = push(head, 9);
        head = push(head, 10);
        head = push(head, 16);
        head = push(head, 15);
 
        int prod = prodOfPrime(head);
        System.out.println("Product of Prime Nodes: " + prod);
    }
}
 
// This code is contributed by Vivekkumar Singh

Python3

# Python3 implementation to product all
# prime nodes from the doubly
# linked list
 
# Node of the doubly linked list
class Node:
     
    def __init__(self, data):
        self.data = data
        self.prev = None
        self.next = None
 
# function to insert a node at the beginning
# of the Doubly Linked List
def push(head_ref, new_data):
 
    # allocate node
    new_node = Node(0)
 
    # put in the data
    new_node.data = new_data
 
    # since we are multiplying at the beginning,
    # prev is always None
    new_node.prev = None
 
    # link the old list off the new node
    new_node.next = (head_ref)
 
    # change prev of head node to new node
    if ((head_ref) != None):
        (head_ref).prev = new_node
 
    # move the head to point to the new node
    (head_ref) = new_node
    return head_ref
 
# Function to check if a number is prime
def isPrime(n):
 
    # Corner cases
    if (n <= 1):
        return False
    if (n <= 3):
        return True
 
    # This is checked so that we can skip
    # middle five numbers in below loop
    if (n % 2 == 0 or n % 3 == 0):
        return False
    i = 5
    while ( i * i <= n ):
        if (n % i == 0 or n % (i + 2) == 0):
            return False
        i += 6;
 
    return True
 
# function to product all prime nodes
# from the doubly linked list
def prodOfPrime(head_ref):
 
    ptr = head_ref
    next = None
     
    # variable prod = 1 for multiplying nodes value
    prod = 1
     
    # traverse list till last node
    while (ptr != None):
        next = ptr.next
         
        # if number is prime then
        # multiply to product
        if (isPrime(ptr.data)):
            prod = prod * ptr.data
        ptr = next
 
    # return product
    return prod
 
# Driver Code
if __name__ == "__main__":
 
    # start with the empty list
    head = None
 
    # create the doubly linked list
    # 15 <. 16 <. 7 <. 6 <. 17
    head = push(head, 17)
    head = push(head, 6)
    head = push(head, 7)
    head = push(head, 16)
    head = push(head, 15)
    prod = prodOfPrime(head)
 
    print("Product of Prime Nodes : ", prod)
 
# This code is contributed by Arnab Kundu

C#

// C# implementation to product all
// prime nodes from the doubly
// linked list
using System;
 
// Node of the doubly linked list
public class Node
{
    public int data;
    public Node next, prev;
 
    public Node(int d)
    {
        data = d;
        next = null;
        prev = null;
    }
}
 
class DLL
{
    // function to insert a node at the beginning
    // of the Doubly Linked List
    static Node push(Node head, int data)
    {
        Node newNode = new Node(data);
        newNode.next = head;
        newNode.prev = null;
        if (head != null)
            head.prev = newNode;
        head = newNode;
 
        return head;
    }
 
    // Function to check if a number is prime
    static Boolean isPrime(int n)
    {
        // Corner cases
        if (n <= 1)
            return false;
        if (n <= 3)
            return true;
 
        // This is checked so that we can skip
        // middle five numbers in below loop
        if (n % 2 == 0 || n % 3 == 0)
            return false;
 
        for (int i = 5; i * i <= n; i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;
 
        return true;
    }
 
    // function to product all prime nodes
    // from the doubly linked list
    static int prodOfPrime(Node node)
    {
        // variable prod = 1 for multiplying nodes value
        int prod = 1;
         
        // traverse list till last node
        while (node != null)
        {
            // check is node value is Prime
            // if true then multiply to prod
            if (isPrime(node.data))
                prod *= node.data;
            node = node.next;
        }
        // return product
        return prod;
    }
 
    // Driver code
    public static void Main(String []args)
    {
        // Start with empty list
        Node head = null;
 
        // create the doubly linked list
        // 15 <-> 16 <-> 7 <-> 6 <-> 17
        head = push(head, 17);
        head = push(head, 7);
        head = push(head, 6);
        head = push(head, 9);
        head = push(head, 10);
        head = push(head, 16);
        head = push(head, 15);
 
        int prod = prodOfPrime(head);
        Console.WriteLine("Product of Prime Nodes: " + prod);
    }
}
 
// This code is contributed by Arnab Kundu

Javascript

 
// JavaScript implementation to product all
// prime nodes from the doubly
// linked list
 
// Node of the doubly linked list
class Node {
    constructor(val) {
        this.data = val;
        this.prev = null;
        this.next = null;
    }
}
 
 
    // function to insert a node at the beginning
    // of the Doubly Linked List
    function push(head , data) {
    var newNode = new Node(data);
        newNode.next = head;
        newNode.prev = null;
        if (head != null)
            head.prev = newNode;
        head = newNode;
 
        return head;
    }
 
    // Function to check if a number is prime
    function isPrime(n) {
        // Corner cases
        if (n <= 1)
            return false;
        if (n <= 3)
            return true;
 
        // This is checked so that we can skip
        // middle five numbers in below loop
        if (n % 2 == 0 || n % 3 == 0)
            return false;
 
        for (i = 5; i * i <= n; i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;
 
        return true;
    }
 
    // function to product all prime nodes
    // from the doubly linked list
    function prodOfPrime(node) {
        // variable prod = 1 for multiplying nodes value
        var prod = 1;
 
        // traverse list till last node
        while (node != null) {
            // check is node value is Prime
            // if true then multiply to prod
            if (isPrime(node.data))
                prod *= node.data;
            node = node.next;
        }
        // return product
        return prod;
    }
 
    // Driver Program
     
        // Start with empty list
        var head = null;
 
        // create the doubly linked list
        // 15 <-> 16 <-> 7 <-> 6 <-> 17
        head = push(head, 17);
        head = push(head, 7);
        head = push(head, 6);
        head = push(head, 9);
        head = push(head, 10);
        head = push(head, 16);
        head = push(head, 15);
 
        var prod = prodOfPrime(head);
        document.write("Product of Prime Nodes: " + prod);
 
// This code contributed by umadevi9616
 

输出：

Product of Prime Nodes : 119

时间复杂度：O(N)，其中 N 是节点数。