从两个链表中计算乘积等于给定值的对

给定两个不同元素的大小为n1和n2的链表（可以排序或未排序）。给定一个值 X。问题是计算两个列表中乘积等于给定值 x 的所有对。
注意：该对必须具有每个链表中的一个元素。
例子：

Input : list1 = 3->1->5->7
        list2 = 8->2->5->3
        X = 10
Output : 1
The pair is: (5, 2)

Input : list1 = 4->3->5->7->11->2->1
        list2 = 2->3->4->5->6->8-12
        X = 9   
Output : 1
The pair is: (3, 3)

一个简单的方法是使用两个循环从两个链表中选取元素并检查该对的乘积是否等于给定的值 X。计算所有这些对并打印结果。
下面是上述方法的实现：

C++

// C++ program to count all pairs from both the
// linked lists whose product is equal to
// a given value
 
#include 
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 = (struct Node*)malloc(sizeof(struct 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 count all pairs from both the linked lists
// whose product is equal to a given value
int countPairs(struct Node* head1, struct Node* head2, int x)
{
    int count = 0;
 
    struct Node *p1, *p2;
 
    // Traverse the 1st linked list
    for (p1 = head1; p1 != NULL; p1 = p1->next) {
        // for each node of 1st list
        // Traverse the 2nd list
        for (p2 = head2; p2 != NULL; p2 = p2->next) {
            // if sum of pair is equal to 'x'
            // increment count
            if ((p1->data * p2->data) == x)
                count++;
        }
    }
 
    // required count of pairs
    return count;
}
 
// Driver Code
int main()
{
    struct Node* head1 = NULL;
    struct Node* head2 = NULL;
 
    // create linked list1 3->1->5->7
    push(&head1, 7);
    push(&head1, 5);
    push(&head1, 1);
    push(&head1, 3);
 
    // create linked list2 8->2->5->3
    push(&head2, 3);
    push(&head2, 5);
    push(&head2, 2);
    push(&head2, 8);
 
    int x = 10;
 
    cout << "Count = " << countPairs(head1, head2, x);
 
    return 0;
}

Java

// Java program to count all pairs from both the
// linked lists whose product is equal to
// a given value
 
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 count all pairs from both the linked lists
    // whose product is equal to a given value
    static int countPairs(Node head1, Node head2, int x)
    {
        int count = 0;
 
        Node p1, p2;
 
        // Traverse the 1st linked list
        for (p1 = head1; p1 != null; p1 = p1.next)
        {
            // for each node of 1st list
            // Traverse the 2nd list
            for (p2 = head2; p2 != null; p2 = p2.next)
            {
                // if sum of pair is equal to 'x'
                // increment count
                if ((p1.data * p2.data) == x)
                {
                    count++;
                }
            }
        }
 
        // required count of pairs
        return count;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        Node head1 = null;
        Node head2 = null;
 
        // create linked list1 3.1.5.7
        head1 = push(head1, 7);
        head1 = push(head1, 5);
        head1 = push(head1, 1);
        head1 = push(head1, 3);
 
        // create linked list2 8.2.5.3
        head2 = push(head2, 3);
        head2 = push(head2, 5);
        head2 = push(head2, 2);
        head2 = push(head2, 8);
 
        int x = 10;
 
        System.out.print("Count = " + countPairs(head1, head2, x));
    }
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program to count all pairs from both the
# linked lists whose product is equal to
# a given value
  
''' 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 point to the new node '''
    (head_ref) = new_node;
     
    return head_ref
  
# Function to count all pairs from both the linked lists
# whose product is equal to a given value
def countPairs(head1, head2, x):
    count = 0;   
    p1 = head1
     
    # Traverse the 1st linked list
    while p1 != None:       
        p2 = head2
         
        # for each node of 1st list
        # Traverse the 2nd list
        while p2 != None:       
         
            # if sum of pair is equal to 'x'
            # increment count
            if ((p1.data * p2.data) == x):
                count += 1
            p2 = p2.next
        p1 = p1.next
         
    # required count of pairs
    return count;
  
# Driver Code
if __name__=='__main__':
     
    head1 = None;
    head2 = None;
  
    # create linked list1 3.1.5.7
    head1 = push(head1, 7);
    head1 = push(head1, 5);
    head1 = push(head1, 1);
    head1 = push(head1, 3);
  
    # create linked list2 8.2.5.3
    head2 = push(head2, 3);
    head2 = push(head2, 5);
    head2 = push(head2, 2);
    head2 = push(head2, 8);
  
    x = 10;
  
    print("Count = " + str(countPairs(head1, head2, x)))
  
# This code is contributed by rutvik_56

C#

// C# program to count all pairs from both the
// linked lists whose product is equal to
// a given value
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 count all pairs from both the linked lists
    // whose product is equal to a given value
    static int countPairs(Node head1, Node head2, int x)
    {
        int count = 0;
 
        Node p1, p2;
 
        // Traverse the 1st linked list
        for (p1 = head1; p1 != null; p1 = p1.next)
        {
            // for each node of 1st list
            // Traverse the 2nd list
            for (p2 = head2; p2 != null; p2 = p2.next)
            {
                // if sum of pair is equal to 'x'
                // increment count
                if ((p1.data * p2.data) == x)
                {
                    count++;
                }
            }
        }
 
        // required count of pairs
        return count;
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        Node head1 = null;
        Node head2 = null;
 
        // create linked list1 3->1->5->7
        head1 = push(head1, 7);
        head1 = push(head1, 5);
        head1 = push(head1, 1);
        head1 = push(head1, 3);
 
        // create linked list2 8->2->5->3
        head2 = push(head2, 3);
        head2 = push(head2, 5);
        head2 = push(head2, 2);
        head2 = push(head2, 8);
 
        int x = 10;
 
        Console.Write("Count = " + countPairs(head1, head2, x));
    }
}
 
// This code is contributed by Rajput-Ji

Javascript

// javascript program to count all pairs from both the
// linked lists whose product is equal to
// a given value
    /* 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 povar to the new node */
        head_ref = new_node;
        return head_ref;
    }
 
    // Function to count all pairs from both the linked lists
    // whose product is equal to a given value
    function countPairs(head1,  head2 , x) {
        var count = 0;
 
var p1, p2;
 
        // Traverse the 1st linked list
        for (p1 = head1; p1 != null; p1 = p1.next) {
            // for each node of 1st list
            // Traverse the 2nd list
            for (p2 = head2; p2 != null; p2 = p2.next) {
                // if sum of pair is equal to 'x'
                // increment count
                if ((p1.data * p2.data) == x) {
                    count++;
                }
            }
        }
 
        // required count of pairs
        return count;
    }
 
    // Driver Code
     
var head1 = null;
var head2 = null;
 
        // create linked list1 3.1.5.7
        head1 = push(head1, 7);
        head1 = push(head1, 5);
        head1 = push(head1, 1);
        head1 = push(head1, 3);
 
        // create linked list2 8.2.5.3
        head2 = push(head2, 3);
        head2 = push(head2, 5);
        head2 = push(head2, 2);
        head2 = push(head2, 8);
 
        var x = 10;
 
        document.write("Count = " + countPairs(head1, head2, x));
 
// This code contributed by umadevi9616

输出：

Count = 1

时间复杂度： O(N^2)

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