单链接列表的所有偶数和节点的和与乘积

给定一个包含N个节点的单链接列表，任务是从列表中查找其数据值具有偶数位数字和的所有节点的和与乘积。

例子：

Input: 15 -> 16 -> 8 -> 6 -> 13 
Output: 
Sum = 42 
Product = 9360 
Explanation: 
The sum of all digit of number in linked list are: 
15 = 1 + 5 = 6 
16 = 1 + 6 = 7 
8 = 8 
6 = 6 
13 = 1 + 3 = 4 
The list contains 4 Even Digit Sum data values 15, 8, 6 and 13. 
Sum = 15 + 8 + 6 + 13 = 42 
Product = 15 * 8 * 6 * 13 = 9360

Input: 5 -> 3 -> 4 -> 2 -> 9 
Output: 
Sum = 6 
Product = 8 
Explanation: 
The list contains 2 Even Digit Sum data values 4 and 2. 
Sum = 4 + 2 = 6 
Product = 4 * 2 = 8 

方法：想法是遍历给定的链表，并检查当前节点值的所有数字的总和是否为偶数。如果是，则将当前节点值包括到结果总和中，并将结果乘积Else检查下一个节点值。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Node of Linked List
struct Node {
    int data;
    Node* next;
};
 
// Function to insert a node at the
// beginning of the singly Linked List
void push(Node** head_ref, int new_data)
{
    // Allocate new node
    Node* new_node
        = (Node*)malloc(
            sizeof(struct Node));
 
    // Insert the data
    new_node->data = new_data;
 
    // Link old list to the new node
    new_node->next = (*head_ref);
 
    // Move head to point the new node
    (*head_ref) = new_node;
}
 
// Function to find the digit sum
// for a number
int digitSum(int num)
{
    int sum = 0;
    while (num) {
        sum += (num % 10);
        num /= 10;
    }
 
    // Return the sum
    return sum;
}
 
// Function to find the required
// sum and product
void sumAndProduct(Node* head_ref)
{
 
    // Initialise the sum and product
    // to 0 and 1 respectively
    int prod = 1;
    int sum = 0;
 
    Node* ptr = head_ref;
 
    // Traverse the given linked list
    while (ptr != NULL) {
 
        // If current node has even
        // digit sum then include it in
        // resultant sum and product
        if (!(digitSum(ptr->data) & 1)) {
 
            // Find the sum and the product
            prod *= ptr->data;
            sum += ptr->data;
        }
 
        ptr = ptr->next;
    }
 
    // Print the final Sum and Product
    cout << "Sum = " << sum << endl;
    cout << "Product = " << prod;
}
 
// Driver Code
int main()
{
    // Head of the linked list
    Node* head = NULL;
 
    // Create the linked list
    // 15 -> 16 -> 8 -> 6 -> 13
    push(&head, 13);
    push(&head, 6);
    push(&head, 8);
    push(&head, 16);
    push(&head, 15);
 
    // Function call
    sumAndProduct(head);
 
    return 0;
}

Java

// Java program for the above approach
 
class GFG{
 
// Node of Linked List
static class Node {
    int data;
    Node next;
};
 
// Function to insert a node at the
// beginning of the singly Linked List
static Node push(Node head_ref, int new_data)
{
    // Allocate new node
    Node new_node
        = new Node();
 
    // Insert the data
    new_node.data = new_data;
 
    // Link old list to the new node
    new_node.next = head_ref;
 
    // Move head to point the new node
    head_ref = new_node;
    return head_ref;
}
 
// Function to find the digit sum
// for a number
static int digitSum(int num)
{
    int sum = 0;
    while (num > 0) {
        sum += (num % 10);
        num /= 10;
    }
 
    // Return the sum
    return sum;
}
 
// Function to find the required
// sum and product
static void sumAndProduct(Node head_ref)
{
 
    // Initialise the sum and product
    // to 0 and 1 respectively
    int prod = 1;
    int sum = 0;
 
    Node ptr = head_ref;
 
    // Traverse the given linked list
    while (ptr != null) {
 
        // If current node has even
        // digit sum then include it in
        // resultant sum and product
        if ((digitSum(ptr.data) %2 != 1)) {
 
            // Find the sum and the product
            prod *= ptr.data;
            sum += ptr.data;
        }
 
        ptr = ptr.next;
    }
 
    // Print the final Sum and Product
    System.out.print("Sum = " + sum +"\n");
    System.out.print("Product = " + prod);
}
 
// Driver Code
public static void main(String[] args)
{
    // Head of the linked list
    Node head = null;
 
    // Create the linked list
    // 15.16.8.6.13
    head = push(head, 13);
    head = push(head, 6);
    head = push(head, 8);
    head = push(head, 16);
    head = push(head, 15);
 
    // Function call
    sumAndProduct(head);
 
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach
 
# Node of Linked List
class Node:
     
    def __init__(self, x):
         
        self.data = x
        self.next = None
 
# Function to insert a node at the
# beginning of the singly Linked List
def push(head_ref, new_data):
     
    # Insert the data
    new_node = Node(new_data)
 
    # Link old list to the new node
    new_node.next = head_ref
 
    # Move head to pothe new node
    head_ref = new_node
 
    return head_ref
 
# Function to find the digit sum
# for a number
def digitSum(num):
     
    sum = 0
     
    while (num):
        sum += (num % 10)
        num //= 10
 
    # Return the sum
    return sum
 
# Function to find the required
# sum and product
def sumAndProduct(head_ref):
 
    # Initialise the sum and product
    # to 0 and 1 respectively
    prod = 1
    sum = 0
 
    ptr = head_ref
 
    # Traverse the given linked list
    while (ptr != None):
 
        # If current node has even
        # digit sum then include it in
        # resultant sum and product
        if (not (digitSum(ptr.data) & 1)):
 
            # Find the sum and the product
            prod *= ptr.data
            sum += ptr.data
 
        ptr = ptr.next
 
    # Print the final Sum and Product
    print("Sum =", sum)
    print("Product =", prod)
 
# Driver Code
if __name__ == '__main__':
     
    # Head of the linked list
    head = None
 
    # Create the linked list
    # 15 . 16 . 8 . 6 . 13
    head = push(head, 13)
    head = push(head, 6)
    head = push(head, 8)
    head = push(head, 16)
    head = push(head, 15)
 
    # Function call
    sumAndProduct(head)
     
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Node of Linked List
class Node
{
    public int data;
    public Node next;
};
 
// Function to insert a node at the
// beginning of the singly Linked List
static Node push(Node head_ref, int new_data)
{
    // Allocate new node
    Node new_node = new Node();
 
    // Insert the data
    new_node.data = new_data;
 
    // Link old list to the new node
    new_node.next = head_ref;
 
    // Move head to point the new node
    head_ref = new_node;
    return head_ref;
}
 
// Function to find the digit sum
// for a number
static int digitSum(int num)
{
    int sum = 0;
    while (num > 0)
    {
        sum += (num % 10);
        num /= 10;
    }
 
    // Return the sum
    return sum;
}
 
// Function to find the required
// sum and product
static void sumAndProduct(Node head_ref)
{
 
    // Initialise the sum and product
    // to 0 and 1 respectively
    int prod = 1;
    int sum = 0;
 
    Node ptr = head_ref;
 
    // Traverse the given linked list
    while (ptr != null)
    {
 
        // If current node has even
        // digit sum then include it in
        // resultant sum and product
        if ((digitSum(ptr.data) % 2 != 1))
        {
 
            // Find the sum and the product
            prod *= ptr.data;
            sum += ptr.data;
        }
 
        ptr = ptr.next;
    }
 
    // Print the readonly Sum and Product
    Console.Write("Sum = " + sum + "\n");
    Console.Write("Product = " + prod);
}
 
// Driver Code
public static void Main(String[] args)
{
    // Head of the linked list
    Node head = null;
 
    // Create the linked list
    // 15.16.8.6.13
    head = push(head, 13);
    head = push(head, 6);
    head = push(head, 8);
    head = push(head, 16);
    head = push(head, 15);
 
    // Function call
    sumAndProduct(head);
 
}
}
 
// This code is contributed by Rohit_ranjan

输出：

Sum = 42
Product = 9360

时间复杂度： O(N) ，其中N是链表中节点的数量。