计算乘积等于给定值x的有序双链表中的三元组

给定一个不同节点的排序双链表(没有两个节点具有相同的数据)和一个值x。任务是对列表中的三元组进行计数，直到达到给定值x。

例子：

Input: list = 1->2->4->5->6->8->9, x = 8 
Output: 1 
Triplet is (1, 2, 4)

Input: list = 1->2->4->5->6->8->9, x = 120 
Output: 1 
Triplet is (4, 5, 6) 

天真的方法：使用三个嵌套循环生成所有三元组，并检查三元组乘积中的元素是否等于x。

下面是上述方法的实现：

C++

// C++ implementation to count triplets
// in a sorted doubly linked list
// whose product is equal to a given value 'x'
#include 
using namespace std;
 
// structure of node of doubly linked list
struct Node {
    int data;
    struct Node *next, *prev;
};
 
// function to count triplets in a sorted doubly linked list
// whose product is equal to a given value 'x'
int countTriplets(struct Node* head, int x)
{
    struct Node *ptr1, *ptr2, *ptr3;
    int count = 0;
 
    // generate all possible triplets
    for (ptr1 = head; ptr1 != NULL; ptr1 = ptr1->next)
        for (ptr2 = ptr1->next; ptr2 != NULL; ptr2 = ptr2->next)
            for (ptr3 = ptr2->next; ptr3 != NULL; ptr3 = ptr3->next)
 
                // if elements in the current triplet product up to 'x'
                if ((ptr1->data * ptr2->data * ptr3->data) == x)
 
                    // increment count
                    count++;
 
    // required count of triplets
    return count;
}
 
// A utility function to insert a new node at the
// beginning of doubly linked list
void insert(struct Node** head, int data)
{
    // allocate node
    struct Node* temp = new Node();
 
    // put in the data
    temp->data = data;
    temp->next = temp->prev = NULL;
 
    if ((*head) == NULL)
        (*head) = temp;
    else {
        temp->next = *head;
        (*head)->prev = temp;
        (*head) = temp;
    }
}
 
// Driver program to test above
int main()
{
    // start with an empty doubly linked list
    struct Node* head = NULL;
 
    // insert values in sorted order
    insert(&head, 9);
    insert(&head, 8);
    insert(&head, 6);
    insert(&head, 5);
    insert(&head, 4);
    insert(&head, 2);
    insert(&head, 1);
 
    int x = 8;
 
    cout << "Count = "
         << countTriplets(head, x);
    return 0;
}

Java

// Java implementation to count triplets
// in a sorted doubly linked list
// whose sum is equal to a given value 'x'
import java.util.*;
 
// Represents node of a doubly linked list
class Node
{
    public int data;
    public Node prev, next;
    public Node(int val)
    {
        data = val;
        prev = null;
        next = null;
    }
}
 
class GFG
{
 
    // function to count triplets in
    // a sorted doubly linked list
    // whose sum is equal to a given value 'x'
    static int countTriplets(Node head, int x)
    {
        Node ptr1, ptr2, ptr3;
        int count = 0;
 
        // generate all possible triplets
        for (ptr1 = head; ptr1 != null; ptr1 = ptr1.next)
            for (ptr2 = ptr1.next; ptr2 != null; ptr2 = ptr2.next)
                for (ptr3 = ptr2.next; ptr3 != null; ptr3 = ptr3.next)
 
                    // if elements in the current triplet sum up to 'x'
                    if ((ptr1.data * ptr2.data * ptr3.data) == x)
                         
                        // increment count
                        count++;
 
        // required count of triplets
        return count;
    }
 
    // A utility function to insert a new node at the
    // beginning of doubly linked list
    static Node insert(Node head, int val)
    {
        // allocate node
        Node temp = new Node(val);
 
        if (head == null)
            head = temp;
 
        else
        {
            temp.next = head;
            head.prev = temp;
            head = temp;
        }
     
        return head;
    }
 
    // Driver code
    public static void main(String []args)
    {
        // start with an empty doubly linked list
        Node head = null;
         
        // insert values in sorted order
        head = insert(head, 9);
        head = insert(head, 8);
        head = insert(head, 6);
        head = insert(head, 5);
        head = insert(head, 4);
        head = insert(head, 2);
        head = insert(head, 1);
 
        int x = 8;
        System.out.println("count = " + countTriplets(head, x));
    }
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation to count triplets
# in a sorted doubly linked list
# whose sum is equal to a given value 'x'
 
# Represents node of a doubly linked list
class Node:
    data = None
    prev = None
    next_ = None
 
    def __init__(self, val):
        self.data = val
        self.prev = None
        self.next_ = None
 
# function to count triplets in
# a sorted doubly linked list
# whose sum is equal to a given value 'x'
def countTriplets(head, x):
    ptr1, ptr2, ptr3 = Node(0), Node(0), Node(0)
    count = 0
 
    # generate all possible triplets
    ptr1 = head
    while ptr1 is not None:
        ptr2 = ptr1.next_
        while ptr2 is not None:
            ptr3 = ptr2.next_
            while ptr3 is not None:
 
                # if elements in the current
                # triplet sum up to 'x'
                if ptr1.data * ptr2.data * ptr3.data == x:
 
                    # increment count
                    count += 1
                ptr3 = ptr3.next_
            ptr2 = ptr2.next_
        ptr1 = ptr1.next_
 
        # required count of triplets
        return count
 
# A utility function to insert a new node at the
# beginning of doubly linked list
def insert(head, val):
 
    # allocate node
    temp = Node(val)
    if head is None:
        head = temp
    else:
        temp.next_ = head
        head.prev = temp
        head = temp
 
    return head
 
# Driver Code
if __name__ == "__main__":
 
    # start with an empty doubly linked list
    head = Node(0)
 
    # insert values in sorted order
    head = insert(head, 9)
    head = insert(head, 8)
    head = insert(head, 6)
    head = insert(head, 5)
    head = insert(head, 4)
    head = insert(head, 2)
    head = insert(head, 1)
 
    x = 8
    print("count =", countTriplets(head, x))
 
# This code is contributed by
# sanjeev2552

C#

// C# implementation to count triplets
// in a sorted doubly linked list
// whose sum is equal to a given value 'x'
using System;
 
// Represents node of a doubly linked list
public class Node
{
    public int data;
    public Node prev, next;
    public Node(int val)
    {
        data = val;
        prev = null;
        next = null;
    }
}
 
class GFG
{
 
    // function to count triplets in
    // a sorted doubly linked list
    // whose sum is equal to a given value 'x'
    static int countTriplets(Node head, int x)
    {
        Node ptr1, ptr2, ptr3;
        int count = 0;
 
        // generate all possible triplets
        for (ptr1 = head; ptr1 != null; ptr1 = ptr1.next)
            for (ptr2 = ptr1.next; ptr2 != null; ptr2 = ptr2.next)
                for (ptr3 = ptr2.next; ptr3 != null; ptr3 = ptr3.next)
 
                    // if elements in the current triplet sum up to 'x'
                    if ((ptr1.data * ptr2.data * ptr3.data) == x)
                         
                        // increment count
                        count++;
 
        // required count of triplets
        return count;
    }
 
    // A utility function to insert a new node at the
    // beginning of doubly linked list
    static Node insert(Node head, int val)
    {
        // allocate node
        Node temp = new Node(val);
 
        if (head == null)
            head = temp;
 
        else
        {
            temp.next = head;
            head.prev = temp;
            head = temp;
        }
     
        return head;
    }
 
    // Driver code
    public static void Main(String []args)
    {
        // start with an empty doubly linked list
        Node head = null;
         
        // insert values in sorted order
        head = insert(head, 9);
        head = insert(head, 8);
        head = insert(head, 6);
        head = insert(head, 5);
        head = insert(head, 4);
        head = insert(head, 2);
        head = insert(head, 1);
 
        int x = 8;
        Console.WriteLine("count = " + countTriplets(head, x));
    }
}
 
// This code is contributed by Arnab Kundu

输出：

Count = 1

时间复杂度： O(n ^ 3)
辅助空间： O(1)

方法2(哈希)：创建一个哈希表，其中将(键，值)元组表示为(节点数据，节点指针)元组。遍历双向链表，并将每个节点的数据及其指针对(元组)存储在哈希表中。现在，生成每个可能的节点对。对于每对节点，计算p_product(两个节点中数据的乘积)，然后检查哈希表中是否存在(x / p_product)。如果存在，则还要验证该对中的两个节点是否与哈希表中与(x / p_product)关联的节点不同，并最终递增计数。返回(计数/ 3)，因为在上述过程中每个三元组被计数3次。

下面是上述方法的实现：

C++

// C++ implementation to count triplets
// in a sorted doubly linked list
// whose product is equal to a given value 'x'
#include 
using namespace std;
 
// structure of node of doubly linked list
struct Node {
    int data;
    struct Node *next, *prev;
};
 
// function to count triplets in a sorted doubly linked list
// whose product is equal to a given value 'x'
int countTriplets(struct Node* head, int x)
{
    struct Node *ptr, *ptr1, *ptr2;
    int count = 0;
 
    // unordered_map 'um' implemented as hash table
    unordered_map um;
 
    // insert the  tuple in 'um'
    for (ptr = head; ptr != NULL; ptr = ptr->next)
        um[ptr->data] = ptr;
 
    // generate all possible pairs
    for (ptr1 = head; ptr1 != NULL; ptr1 = ptr1->next)
        for (ptr2 = ptr1->next; ptr2 != NULL; ptr2 = ptr2->next) {
 
            // p_product = product of elements in the current pair
            int p_product = (ptr1->data * ptr2->data);
 
            // if 'x/p_product' is present in 'um' and
            // either of the two nodes
            // are not equal to the 'um[x/p_product]' node
            if (um.find(x / p_product) != um.end() && um[x / p_product] != ptr1
                && um[x / p_product] != ptr2)
 
                // increment count
                count++;
        }
 
    // required count of triplets
    // division by 3 as each triplet is counted 3 times
    return (count / 3);
}
 
// A utility function to insert a new node at the
// beginning of doubly linked list
void insert(struct Node** head, int data)
{
    // allocate node
    struct Node* temp = new Node();
 
    // put in the data
    temp->data = data;
    temp->next = temp->prev = NULL;
 
    if ((*head) == NULL)
        (*head) = temp;
    else {
        temp->next = *head;
        (*head)->prev = temp;
        (*head) = temp;
    }
}
 
// Driver program to test above functions
int main()
{
    // start with an empty doubly linked list
    struct Node* head = NULL;
 
    // insert values in sorted order
    insert(&head, 9);
    insert(&head, 8);
    insert(&head, 6);
    insert(&head, 5);
    insert(&head, 4);
    insert(&head, 2);
    insert(&head, 1);
 
    int x = 8;
 
    cout << "Count = "
         << countTriplets(head, x);
    return 0;
}

Java

// Java implementation to count triplets
// in a sorted doubly linked list whose
// product is equal to a given value 'x'
import java.io.*;
import java.util.*;
 
// Structure of node of doubly linked list
class Node
{
    int data;
    Node next, prev;
}
 
class GFG{
     
// Function to count triplets in a sorted
// doubly linked list whose product is
// equal to a given value 'x'
static int countTriplets(Node head, int x)
{
    Node ptr, ptr1, ptr2;
    int count = 0;
     
    // Unordered_map 'um' implemented
    // as hash table
    Map um = new HashMap();
                                
    // Insert the 
    // tuple in 'um'
    for(ptr = head; ptr != null; ptr = ptr.next)
    {
        um.put(ptr.data, ptr);
    }
     
    // Generate all possible pairs
    for(ptr1 = head;
        ptr1 != null;
        ptr1 = ptr1.next)
    {
        for(ptr2 = ptr1.next;
            ptr2 != null;
            ptr2 = ptr2.next)
        {
             
            // p_product = product of elements
            // in the current pair
            int p_product = (ptr1.data * ptr2.data);
             
            // If 'x/p_product' is present in 'um' and
            // either of the two nodes are not equal
            // to the 'um[x/p_product]' node
            if (um.containsKey(x / p_product) &&
                um.get(x / p_product) != ptr1 &&
                um.get(x / p_product) != ptr2)
            {
                 
                // Increment count
                count++;
            }
        }
    }
     
    // Required count of triplets
    // division by 3 as each triplet
    // is counted 3 times
    return (count / 3);
}
 
// A utility function to insert a new
// node at the beginning of doubly linked list
static Node insert(Node head, int data)
{
     
    // Allocate node
    Node temp = new Node();
     
    // Put in the data
    temp.data = data;
    temp.next = temp.prev = null;
     
    if (head == null)
    {
        head = temp;
    }
    else
    {
        temp.next = head;
        head.prev = temp;
        head = temp;
    }
    return head;
}
 
// Driver code
public static void main(String[] args)
{
     
    // Start with an empty doubly linked list
    Node head = null;
     
    // Insert values in sorted order
    head = insert(head, 9);
    head = insert(head, 8);
    head = insert(head, 6);
    head = insert(head, 5);
    head = insert(head, 4);
    head = insert(head, 2);
    head = insert(head, 1);
     
    int x = 8;
     
    System.out.println("Count = " +
                       countTriplets(head, x));
}
}
 
// This code is contributed by avanitrachhadiya2155

Python3

# Python3 implementation to
# count triplets in a sorted
# doubly linked list whose
# product is equal to a given
# value 'x'
from math import ceil
 
# structure of node of doubly
# linked list
class Node:
   
    def __init__(self, x):
       
        self.data = x
        self.next = None
        self.prev = None
 
# function to count triplets
# in a sorted doubly linked
# list whose product is equal
# to a given value 'x'
def countTriplets(head, x):
 
    ptr, ptr1, ptr2 = None, None, None
    count = 0
 
    # unordered_map 'um' implemented
    # as hash table
    um = {}
 
    # insert the  tuple in 'um'
    ptr = head
    while ptr != None:
        um[ptr.data] = ptr
        ptr = ptr.next
 
    # generate all possible pairs
    ptr1 = head
     
    while ptr1 != None:
        ptr2 = ptr1.next
        while ptr2 != None:
 
            # p_product = product of
            # elements in the current
            # pair
            p_product = (ptr1.data *
                         ptr2.data)
 
            # if 'x/p_product' is present
            # in 'um' and either of the
            # two nodes are not equal to
            # the 'um[x/p_product]' node
            if ((x / p_product) in um and
               (x / p_product) != ptr1 and
                um[x / p_product] != ptr2):
 
                # increment count
                count += 1
            ptr2 = ptr2.next
        ptr1 = ptr1.next
 
    # required count of triplets
    # division by 3 as each triplet
    # is counted 3 times
    return (count // 3)
 
# A utility function to insert a
# new node at the beginning of
# doubly linked list
def insert(head, data):
   
    # allocate node
    temp = Node(data)
 
    # put in the data
    temp.data = data
    temp.next = temp.prev = None
 
    if (head == None):
        head = temp
    else:
        temp.next = head
        head.prev = temp
        head = temp
    return head
 
# Driver code
if __name__ == '__main__':
   
    # start with an empty
    # doubly linked list
    head = None
 
    # insert values in sorted
    # order
    head = insert(head, 9)
    head = insert(head, 8)
    head = insert(head, 6)
    head = insert(head, 5)
    head = insert(head, 4)
    head = insert(head, 2)
    head = insert(head, 1)
 
    x = 8
    print("Count =",
           countTriplets(head, x))
 
# This code is contributed by Mohit Kumar 29

C＃

// C# implementation to count triplets
// in a sorted doubly linked list whose
// product is equal to a given value 'x'
using System;
using System.Collections.Generic;
 
// Structure of node of doubly linked list
class Node
{
    public int data;
    public Node next, prev;
}
 
class GFG
{
 
// Function to count triplets in a sorted
// doubly linked list whose product is
// equal to a given value 'x'
static int countTriplets(Node head, int x)
{
    Node ptr, ptr1, ptr2;
    int count = 0;
      
    // Unordered_map 'um' implemented
    // as hash table
    Dictionary um = new Dictionary();
     
    // Insert the 
    // tuple in 'um'
    for(ptr = head; ptr != null; ptr = ptr.next)
    {
        um.Add(ptr.data, ptr);
    }
     // Generate all possible pairs
    for(ptr1 = head;
        ptr1 != null;
        ptr1 = ptr1.next)
    {
        for(ptr2 = ptr1.next;
            ptr2 != null;
            ptr2 = ptr2.next)
        {
              
            // p_product = product of elements
            // in the current pair
            int p_product = (ptr1.data * ptr2.data);
              
            // If 'x/p_product' is present in 'um' and
            // either of the two nodes are not equal
            // to the 'um[x/p_product]' node
            if (um.ContainsKey(x / p_product) &&
                um[x / p_product] != ptr1 &&
                um[x / p_product] != ptr2)
            {
                  
                // Increment count
                count++;
            }
        }
    }
      
    // Required count of triplets
    // division by 3 as each triplet
    // is counted 3 times
    return (count / 3);
}
 
// A utility function to insert a new
// node at the beginning of doubly linked list
static Node insert(Node head, int data)
{
      
    // Allocate node
    Node temp = new Node();
      
    // Put in the data
    temp.data = data;
    temp.next = temp.prev = null;   
    if (head == null)
    {
        head = temp;
    }
    else
    {
        temp.next = head;
        head.prev = temp;
        head = temp;
    }
    return head;
}
  
// Driver code    
static public void Main ()
{
   
    // Start with an empty doubly linked list
    Node head = null;
      
    // Insert values in sorted order
    head = insert(head, 9);
    head = insert(head, 8);
    head = insert(head, 6);
    head = insert(head, 5);
    head = insert(head, 4);
    head = insert(head, 2);
    head = insert(head, 1);   
    int x = 8;
    Console.WriteLine("Count = " + countTriplets(head, x));
}
}
 
// This code is contributed by rag2127

输出：

Count = 1

时间复杂度： O(n ^ 2)
辅助空间： O(n)

方法3(使用两个指针)：从左到右遍历双向链表。对于遍历期间的每个当前节点，初始化两个指针：first =指向当前节点旁边的节点的指针和last =指向列表中最后一个节点的指针。现在，对列表中从第一个指针到最后一个指针的对进行计数，直到达到值(x /当前节点的数据)为止(本文中描述的算法)。将此计数添加到三元组的total_count中。指向最后一个节点的指针只能在开始时找到一次。

下面是上述方法的实现：

C++

// C++ implementation to count triplets
// in a sorted doubly linked list
// whose product is equal to a given value 'x'
#include 
using namespace std;
 
// structure of node of doubly linked list
struct Node {
    int data;
    struct Node *next, *prev;
};
 
// function to count pairs whose product equal to given 'value'
int countPairs(struct Node* first, struct Node* second, int value)
{
    int count = 0;
 
    // The loop terminates when either of two pointers
    // become NULL, or they cross each other (second->next
    // == first), or they become same (first == second)
    while (first != NULL && second != NULL && first != second
           && second->next != first) {
 
        // pair found
        if ((first->data * second->data) == value) {
 
            // increment count
            count++;
 
            // move first in forward direction
            first = first->next;
 
            // move second in backward direction
            second = second->prev;
        }
 
        // if product is greater than 'value'
        // move second in backward direction
        else if ((first->data * second->data) > value)
            second = second->prev;
 
        // else move first in forward direction
        else
            first = first->next;
    }
 
    // required count of pairs
    return count;
}
 
// function to count triplets in a sorted doubly linked list
// whose product is equal to a given value 'x'
int countTriplets(struct Node* head, int x)
{
    // if list is empty
    if (head == NULL)
        return 0;
 
    struct Node *current, *first, *last;
    int count = 0;
 
    // get pointer to the last node of
    // the doubly linked list
    last = head;
    while (last->next != NULL)
        last = last->next;
 
    // traversing the doubly linked list
    for (current = head; current != NULL; current = current->next) {
 
        // for each current node
        first = current->next;
 
        // count pairs with product(x / current->data) in the range
        // first to last and add it to the 'count' of triplets
        count += countPairs(first, last, x / current->data);
    }
 
    // required count of triplets
    return count;
}
 
// A utility function to insert a new node at the
// beginning of doubly linked list
void insert(struct Node** head, int data)
{
    // allocate node
    struct Node* temp = new Node();
 
    // put in the data
    temp->data = data;
    temp->next = temp->prev = NULL;
 
    if ((*head) == NULL)
        (*head) = temp;
    else {
        temp->next = *head;
        (*head)->prev = temp;
        (*head) = temp;
    }
}
 
// Driver program to test above
int main()
{
    // start with an empty doubly linked list
    struct Node* head = NULL;
 
    // insert values in sorted order
    insert(&head, 9);
    insert(&head, 8);
    insert(&head, 6);
    insert(&head, 5);
    insert(&head, 4);
    insert(&head, 2);
    insert(&head, 1);
 
    int x = 8;
 
    cout << "Count = "
         << countTriplets(head, x);
    return 0;
}

Java

// Java implementation to count triplets
// in a sorted doubly linked list
// whose product is equal to a given value 'x'
import java.util.*;
 
class GFG
{
     
// structure of node of doubly linked list
static class Node
{
    int data;
    Node next, prev;
};
 
// function to count pairs whose product
// equal to given 'value'
static int countPairs(Node first, Node second,
                      int value)
{
    int count = 0;
 
    // The loop terminates when either of two pointers
    // become null, or they cross each other (second.next
    // == first), or they become same (first == second)
    while (first != null && second != null &&
            first != second && second.next != first)
    {
 
        // pair found
        if ((first.data * second.data) == value)
        {
 
            // increment count
            count++;
 
            // move first in forward direction
            first = first.next;
 
            // move second in backward direction
            second = second.prev;
        }
 
        // if product is greater than 'value'
        // move second in backward direction
        else if ((first.data * second.data) > value)
            second = second.prev;
 
        // else move first in forward direction
        else
            first = first.next;
    }
 
    // required count of pairs
    return count;
}
 
// function to count triplets in
// a sorted doubly linked list
// whose product is equal to a given value 'x'
static int countTriplets(Node head, int x)
{
    // if list is empty
    if (head == null)
        return 0;
 
    Node current, first, last;
    int count = 0;
 
    // get pointer to the last node of
    // the doubly linked list
    last = head;
    while (last.next != null)
        last = last.next;
 
    // traversing the doubly linked list
    for (current = head; current != null;
        current = current.next)
    {
 
        // for each current node
        first = current.next;
 
        // count pairs with product(x / current.data)
        // in the range first to last and
        // add it to the 'count' of triplets
        count += countPairs(first, last, x / current.data);
    }
 
    // required count of triplets
    return count;
}
 
// A utility function to insert a new node at the
// beginning of doubly linked list
static Node insert(Node head, int data)
{
    // allocate node
    Node temp = new Node();
 
    // put in the data
    temp.data = data;
    temp.next = temp.prev = null;
 
    if ((head) == null)
        (head) = temp;
    else
    {
        temp.next = head;
        (head).prev = temp;
        (head) = temp;
    }
    return head;
}
 
// Driver code
public static void main(String args[])
{
    // start with an empty doubly linked list
    Node head = null;
 
    // insert values in sorted order
    head = insert(head, 9);
    head = insert(head, 8);
    head = insert(head, 6);
    head = insert(head, 5);
    head = insert(head, 4);
    head = insert(head, 2);
    head = insert(head, 1);
 
    int x = 8;
 
    System.out.println( "Count = "+ countTriplets(head, x));
}
}
 
// This code is contributed by Arnab Kundu

Python3

# Python3 implementation to count triplets
# in a sorted doubly linked list whose
# product is equal to a given value 'x'
      
# Structure of node of doubly linked list
class Node:
     
    def __init__(self, data):
         
        self.data = data
        self.next = None
        self.prev = None
 
# Function to count pairs whose product
# equal to given 'value'
def countPairs(first, second, value):
     
    count = 0
  
    # The loop terminates when either of two pointers
    # become None, or they cross each other (second.next
    # == first), or they become same (first == second)
    while (first != None and second != None and
            first != second and second.next != first):
 
        # Pair found
        if ((first.data * second.data) == value):
             
            # Increment count
            count += 1
  
            # Move first in forward direction
            first = first.next
  
            # Move second in backward direction
            second = second.prev
  
        # If product is greater than 'value'
        # move second in backward direction
        elif ((first.data * second.data) > value):
            second = second.prev
  
        # Else move first in forward direction
        else:
            first = first.next
  
    # Required count of pairs
    return count
  
# Function to count triplets in a sorted
# doubly linked list whose product is
# equal to a given value 'x'
def countTriplets(head, x):
 
    # If list is empty
    if (head == None):
        return 0
  
    count = 0
  
    # Get pointer to the last node of
    # the doubly linked list
    last = head
     
    while (last.next != None):
        last = last.next
         
    current = head
     
    # Traversing the doubly linked list
    while current != None:
         
        # For each current node
        first = current.next
  
        # Count pairs with product(x / current.data)
        # in the range first to last and
        # add it to the 'count' of triplets
        count += countPairs(first, last,
                            x // current.data)
        current = current.next
  
    # Required count of triplets
    return count
 
# A utility function to insert a new node
# at the beginning of doubly linked list
def insert(head, data):
 
    # Allocate node
    temp = Node(data)
  
    # Put in the data
    temp.data = data
    temp.next = temp.prev = None
  
    if ((head) == None):
        (head) = temp
    else:
        temp.next = head
        (head).prev = temp
        (head) = temp
     
    return head
 
# Driver code
if __name__=='__main__':
     
    # Start with an empty doubly linked list
    head = None
  
    # Insert values in sorted order
    head = insert(head, 9)
    head = insert(head, 8)
    head = insert(head, 6)
    head = insert(head, 5)
    head = insert(head, 4)
    head = insert(head, 2)
    head = insert(head, 1)
  
    x = 8
  
    print( "Count = " + str(countTriplets(head, x)))
 
# This code is contributed by rutvik_56

C＃

// C# implementation to count triplets
// in a sorted doubly linked list
// whose product is equal to a given value 'x'
using System;
 
class GFG
{
      
// structure of node of doubly linked list
class Node
{
    public int data;
    public Node next, prev;
};
  
// function to count pairs whose product
// equal to given 'value'
static int countPairs(Node first, Node second,
                      int value)
{
    int count = 0;
  
    // The loop terminates when either of two pointers
    // become null, or they cross each other (second.next
    // == first), or they become same (first == second)
    while (first != null && second != null &&
            first != second && second.next != first)
    {
  
        // pair found
        if ((first.data * second.data) == value)
        {
  
            // increment count
            count++;
  
            // move first in forward direction
            first = first.next;
  
            // move second in backward direction
            second = second.prev;
        }
  
        // if product is greater than 'value'
        // move second in backward direction
        else if ((first.data * second.data) > value)
            second = second.prev;
  
        // else move first in forward direction
        else
            first = first.next;
    }
  
    // required count of pairs
    return count;
}
  
// function to count triplets in
// a sorted doubly linked list
// whose product is equal to a given value 'x'
static int countTriplets(Node head, int x)
{
    // if list is empty
    if (head == null)
        return 0;
  
    Node current, first, last;
    int count = 0;
  
    // get pointer to the last node of
    // the doubly linked list
    last = head;
    while (last.next != null)
        last = last.next;
  
    // traversing the doubly linked list
    for (current = head; current != null;
        current = current.next)
    {
  
        // for each current node
        first = current.next;
  
        // count pairs with product(x / current.data)
        // in the range first to last and
        // add it to the 'count' of triplets
        count += countPairs(first, last, x / current.data);
    }
  
    // required count of triplets
    return count;
}
  
// A utility function to insert a new node at the
// beginning of doubly linked list
static Node insert(Node head, int data)
{
    // allocate node
    Node temp = new Node();
  
    // put in the data
    temp.data = data;
    temp.next = temp.prev = null;
  
    if ((head) == null)
        (head) = temp;
    else
    {
        temp.next = head;
        (head).prev = temp;
        (head) = temp;
    }
    return head;
}
  
// Driver code
public static void Main(String []args)
{
    // start with an empty doubly linked list
    Node head = null;
  
    // insert values in sorted order
    head = insert(head, 9);
    head = insert(head, 8);
    head = insert(head, 6);
    head = insert(head, 5);
    head = insert(head, 4);
    head = insert(head, 2);
    head = insert(head, 1);
  
    int x = 8;
  
    Console.WriteLine( "Count = "+ countTriplets(head, x));
}
}
 
// This code is contributed by Rajput-Ji

输出：

Count = 1

时间复杂度： O(n ^ 2)
辅助空间： O(1)