查找给定链表的最后 N 个节点的乘积

给定一个链表和一个数 N。求链表最后 n 个节点的乘积。
约束： 0 <= N <= 链表中的节点数。

例子：

Input : List = 10->6->8->4->12, N = 2
Output : 48
Explanation : Product of last two nodes:
              12 * 4 = 48

Input : List = 15->7->9->5->16->14, N = 4
Output : 10080
Explanation : Product of last four nodes:
              9 * 5 * 16 * 14 = 10080

方法一：（使用自定义栈的迭代方式）从左到右遍历节点。遍历时将节点推送到用户定义的堆栈。然后从堆栈中弹出前 n 个值并找到它们的乘积。
下面是上述方法的实现：

C++

// C++ implementation to find the product of last
// 'n' nodes of the Linked List
#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 = 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;
}
 
// utility function to find the product of last 'n' nodes
int productOfLastN_NodesUtil(struct Node* head, int n)
{
    // if n == 0
    if (n <= 0)
        return 0;
 
    stack st;
    int prod = 1;
 
    // traverses the list from left to right
    while (head != NULL) {
 
        // push the node's data onto the stack 'st'
        st.push(head->data);
 
        // move to next node
        head = head->next;
    }
 
    // pop 'n' nodes from 'st' and
    // add them
    while (n--) {
        prod *= st.top();
        st.pop();
    }
 
    // required product
    return prod;
}
 
// Driver program to test above
int main()
{
    struct Node* head = NULL;
 
    // create linked list 10->6->8->4->12
    push(&head, 12);
    push(&head, 4);
    push(&head, 8);
    push(&head, 6);
    push(&head, 10);
 
    int n = 2;
    cout << productOfLastN_NodesUtil(head, n);
    return 0;
}

Java

// Java implementation to find the product
// of last 'n' nodes of the Linked List
import java.util.*;
 
class GFG
{
 
/* A Linked list node */
static class Node
{
    int data;
    Node next;
};
static Node head;
 
// function to insert a node at the
// beginning of the linked list
static void 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;
    head = head_ref;
}
 
// utility function to find the product
// of last 'n' nodes
static int productOfLastN_NodesUtil(Node head,
                                        int n)
{
    // if n == 0
    if (n <= 0)
        return 0;
 
    Stack st = new Stack();
    int prod = 1;
 
    // traverses the list from left to right
    while (head != null)
    {
 
        // push the node's data
        // onto the stack 'st'
        st.push(head.data);
 
        // move to next node
        head = head.next;
    }
 
    // pop 'n' nodes from 'st' and
    // add them
    while (n-- >0)
    {
        prod *= st.peek();
        st.pop();
    }
 
    // required product
    return prod;
}
 
// Driver Code
public static void main(String[] args)
{
    head = null;
 
    // create linked list 10->6->8->4->12
    push(head, 12);
    push(head, 4);
    push(head, 8);
    push(head, 6);
    push(head, 10);
 
    int n = 2;
    System.out.println(productOfLastN_NodesUtil(head, n));
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation to find the product
# of last 'n' nodes of the Linked List
 
# Link list node
class Node:
     
    def __init__(self, data):
        self.data = data
        self.next = next
         
head = None
 
# function to insert a node at the
# beginning of the linked list
def push(head_ref, new_data):
 
    global head
     
    # allocate node
    new_node = Node(0)
 
    # 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
    head = head_ref
 
# utility function to find the product
# of last 'n' nodes
def productOfLastN_NodesUtil(head, n):
 
    # if n == 0
    if (n <= 0):
        return 0
 
    st = []
    prod = 1
 
    # traverses the list from left to right
    while (head != None) :
     
        # push the node's data
        # onto the stack 'st'
        st.append(head.data)
 
        # move to next node
        head = head.next
     
    # pop 'n' nodes from 'st' and
    # add them
    while (n > 0) :
        n = n - 1
        prod *= st[-1]
        st.pop()
     
    # required product
    return prod
 
# Driver Code
 
head = None
 
# create linked list 10->6->8->4->12
push(head, 12)
push(head, 4)
push(head, 8)
push(head, 6)
push(head, 10)
 
n = 2
print(productOfLastN_NodesUtil(head, n))
 
# This code is contributed by Arnab Kundu

C#

// C# implementation to find the product
// of last 'n' nodes of the Linked List
using System;
 
class GFG
{
 
/* A Linked list node */
public 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;
}
 
// utility function to find the product
// of last 'n' nodes
static int productOfLastN_NodesUtil(Node head,
                                    int n)
{
    // if n == 0
    if (n <= 0)
        return 0;
 
    int prod = 1, len = 0;
    Node temp = head;
 
    // calculate the length of the linked list
    while (temp != null)
    {
        len++;
        temp = temp.next;
    }
 
    // count of first (len - n) nodes
    int c = len - n;
    temp = head;
 
    // just traverse the 1st 'c' nodes
    while (temp != null && c-- >0)
 
        // move to next node
        temp = temp.next;
 
    // now traverse the last 'n' nodes
    // and add them
    while (temp != null)
    {
        // accumulate node's data to sum
        prod *= temp.data;
 
        // move to next node
        temp = temp.next;
    }
 
    // required product
    return prod;
}
 
// Driver Code
public static void Main(String[] args)
{
    Node head = null;
 
    // create 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 n = 2;
    Console.Write(productOfLastN_NodesUtil(head, n));
}
}
 
// This code is contributed by Rajput-Ji

Javascript

// javascript implementation to find the product
// of last 'n' nodes of the Linked List
 
    /* A Linked list node */
class Node {
    constructor(val) {
        this.data = val;
        this.next = null;
    }
}
 
    var head;
 
    // 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 point to the new node */
        (head_ref) = new_node;
        head = head_ref;
    }
 
    // utility function to find the product
    // of last 'n' nodes
    function productOfLastN_NodesUtil(head , n) {
        // if n == 0
        if (n <= 0)
            return 0;
 
        var st = [];
        var prod = 1;
 
        // traverses the list from left to right
        while (head != null) {
 
            // push the node's data
            // onto the stack 'st'
            st.push(head.data);
 
            // move to next node
            head = head.next;
        }
 
        // pop 'n' nodes from 'st' and
        // add them
        while (n-- > 0) {
            prod *= st.pop();
             
        }
 
        // required product
        return prod;
    }
 
    // Driver Code
     
        head = null;
 
        // create linked list 10->6->8->4->12
        push(head, 12);
        push(head, 4);
        push(head, 8);
        push(head, 6);
        push(head, 10);
 
        var n = 2;
        document.write(productOfLastN_NodesUtil(head, n));
 
// This code contributed by aashish1995

输出：

48

时间复杂度： O(n)

方法二：（使用系统调用栈的递归方式）递归遍历链表直到结束。现在在从函数调用返回期间，将最后 n 个节点相乘。乘积可以累积在通过引用传递给函数或某个全局变量的某个变量中。
下面是上述方法的实现：

C++

// C++ implementation to find the product of
// last 'n' nodes of the Linked List
#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 = 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 recursively find the product of last
// 'n' nodes of the given linked list
void productOfLastN_Nodes(struct Node* head, int* n,
                          int* prod)
{
    // if head = NULL
    if (!head)
        return;
 
    // recursively traverse the remaining nodes
    productOfLastN_Nodes(head->next, n, prod);
 
    // if node count 'n' is greater than 0
    if (*n > 0) {
 
        // accumulate sum
        *prod = *prod * head->data;
 
        // reduce node count 'n' by 1
        --*n;
    }
}
 
// utility function to find the product of last 'n' nodes
int productOfLastN_NodesUtil(struct Node* head, int n)
{
    // if n == 0
    if (n <= 0)
        return 0;
 
    int prod = 1;
 
    // find the sum of last 'n' nodes
    productOfLastN_Nodes(head, &n, &prod);
 
    // required product
    return prod;
}
 
// Driver program to test above
int main()
{
    struct Node* head = NULL;
 
    // create linked list 10->6->8->4->12
    push(&head, 12);
    push(&head, 4);
    push(&head, 8);
    push(&head, 6);
    push(&head, 10);
 
    int n = 2;
    cout << productOfLastN_NodesUtil(head, n);
    return 0;
}

Java

// Java implementation to find the product of
// last 'n' nodes of the Linked List
 
 
class GFG{
 
static int n, prod;
/* 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 recursively find the product of last
// 'n' nodes of the given linked list
static void productOfLastN_Nodes(Node head)
{
    // if head = null
    if (head==null)
        return;
  
    // recursively traverse the remaining nodes
    productOfLastN_Nodes(head.next);
  
    // if node count 'n' is greater than 0
    if (n > 0) {
  
        // accumulate sum
        prod = prod * head.data;
  
        // reduce node count 'n' by 1
        --n;
    }
}
  
// utility function to find the product of last 'n' nodes
static int productOfLastN_NodesUtil(Node head)
{
    // if n == 0
    if (n <= 0)
     return 0;
  
    prod = 1;
  
    // find the sum of last 'n' nodes
    productOfLastN_Nodes(head);
  
    // required product
    return prod;
}
  
// Driver program to test above
public static void main(String[] args)
{
    Node head = null;
  
 // create 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);
  
    n = 2;
    System.out.print(productOfLastN_NodesUtil(head));
}
}
 
//This code is contributed by 29AjayKumar

Python3

# Python implementation to find the product of
# last 'n' Nodes of the Linked List
 
n, prod = 0, 0;
 
''' A Linked list Node '''
 
class Node:
 
    def __init__(self, data):
        self.data = data
        self.next = next
 
# function to insert a Node at the
# beginning of the linked list
def push(head_ref, new_data):
     
    ''' allocate Node '''
    new_Node = Node(0);
 
    ''' 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 recursively find the product of last
# 'n' Nodes of the given linked list
def productOfLastN_Nodes(head):
    global n, prod;
     
    # if head = None
    if (head == None):
        return;
 
    # recursively traverse the remaining Nodes
    productOfLastN_Nodes(head.next);
 
    # if Node count 'n' is greater than 0
    if (n > 0):
         
        # accumulate sum
        prod = prod * head.data;
 
        # reduce Node count 'n' by 1
        n -= 1;
 
# utility function to find the product of last 'n' Nodes
def productOfLastN_NodesUtil(head):
    global n,prod;
     
    # if n == 0
    if (n <= 0):
        return 0;
 
    prod = 1;
 
    # find the sum of last 'n' Nodes
    productOfLastN_Nodes(head);
     
    # required product
    return prod;
 
# Driver program to test above
if __name__ == '__main__':
    head = None;
    n = 2;
     
    # create 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);
 
    print(productOfLastN_NodesUtil(head));
 
# This code is contributed by 29AjayKumar

C＃

// C# implementation to find the product of
// last 'n' nodes of the Linked List
  
using System;
 
public class GFG{
  
static int n, prod;
/* A Linked list node */
public 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 recursively find the product of last
// 'n' nodes of the given linked list
static void productOfLastN_Nodes(Node head)
{
    // if head = null
    if (head==null)
        return;
   
    // recursively traverse the remaining nodes
    productOfLastN_Nodes(head.next);
   
    // if node count 'n' is greater than 0
    if (n > 0) {
   
        // accumulate sum
        prod = prod * head.data;
   
        // reduce node count 'n' by 1
        --n;
    }
}
   
// utility function to find the product of last 'n' nodes
static int productOfLastN_NodesUtil(Node head)
{
    // if n == 0
    if (n <= 0)
     return 0;
   
    prod = 1;
   
    // find the sum of last 'n' nodes
    productOfLastN_Nodes(head);
   
    // required product
    return prod;
}
   
// Driver program to test above
public static void Main(String[] args)
{
    Node head = null;
   
 // create 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);
   
    n = 2;
    Console.Write(productOfLastN_NodesUtil(head));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// JavaScript implementation to find the product of
// last 'n' nodes of the Linked List
    var n, prod;
 
    /* 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 point to the new node */
        head_ref = new_node;
        return head_ref;
    }
 
    // Function to recursively find the product of last
    // 'n' nodes of the given linked list
    function productOfLastN_Nodes(head) {
        // if head = null
        if (head == null)
            return;
 
        // recursively traverse the remaining nodes
        productOfLastN_Nodes(head.next);
 
        // if node count 'n' is greater than 0
        if (n > 0) {
 
            // accumulate sum
            prod = prod * head.data;
 
            // reduce node count 'n' by 1
            --n;
        }
    }
 
    // utility function to find the product of last 'n' nodes
    function productOfLastN_NodesUtil(head) {
        // if n == 0
        if (n <= 0)
            return 0;
 
        prod = 1;
 
        // find the sum of last 'n' nodes
        productOfLastN_Nodes(head);
 
        // required product
        return prod;
    }
 
    // Driver program to test above
     
        var head = null;
 
        // create 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);
 
        n = 2;
        document.write(productOfLastN_NodesUtil(head));
 
// This code contributed by gauravrajput1
 

输出：

48

时间复杂度： O(n)

方法3（反转链表） ：

以下是步骤：

1. 反转给定的链表。
2. 遍历反向链表的前 n 个节点。
3. 遍历时将它们相乘。
4. 将链表反转回其原始顺序。
5. 退回产品。

下面是上述方法的实现：

C++

// C++ implementation to find the product of last
// 'n' nodes of the Linked List
#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 = 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;
}
 
void reverseList(struct Node** head_ref)
{
    struct Node *current, *prev, *next;
    current = *head_ref;
    prev = NULL;
 
    while (current != NULL) {
        next = current->next;
        current->next = prev;
        prev = current;
        current = next;
    }
 
    *head_ref = prev;
}
 
// utility function to find the product of last 'n' nodes
int productOfLastN_NodesUtil(struct Node* head, int n)
{
    // if n == 0
    if (n <= 0)
        return 0;
 
    // reverse the linked list
    reverseList(&head);
 
    int prod = 1;
    struct Node* current = head;
 
    // traverse the 1st 'n' nodes of the reversed
    // linked list and product them
    while (current != NULL && n--) {
 
        // accumulate node's data to 'sum'
        prod *= current->data;
 
        // move to next node
        current = current->next;
    }
 
    // reverse back the linked list
    reverseList(&head);
 
    // required product
    return prod;
}
 
// Driver program to test above
int main()
{
    struct Node* head = NULL;
 
    // create linked list 10->6->8->4->12
    push(&head, 12);
    push(&head, 4);
    push(&head, 8);
    push(&head, 6);
    push(&head, 10);
 
    int n = 2;
    cout << productOfLastN_NodesUtil(head, n);
    return 0;
}

Java

// Java implementation to find the product of last
// 'n' nodes of the Linked List
 
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;
}
 
static Node reverseList(Node head_ref)
{
    Node current, prev, next;
    current = head_ref;
    prev = null;
 
    while (current != null)
    {
        next = current.next;
        current.next = prev;
        prev = current;
        current = next;
    }
 
    head_ref = prev;
    return head_ref;
}
 
// utility function to find the product of last 'n' nodes
static int productOfLastN_NodesUtil(Node head, int n)
{
    // if n == 0
    if (n <= 0)
        return 0;
 
    // reverse the linked list
    head = reverseList(head);
 
    int prod = 1;
    Node current = head;
 
    // traverse the 1st 'n' nodes of the reversed
    // linked list and product them
    while (current != null && n-- >0)
    {
 
        // accumulate node's data to 'sum'
        prod *= current.data;
 
        // move to next node
        current = current.next;
    }
 
    // reverse back the linked list
    head = reverseList(head);
 
    // required product
    return prod;
}
 
// Driver program to test above
public static void main(String[] args)
{
    Node head = null;
 
    // create 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 n = 2;
    System.out.print(productOfLastN_NodesUtil(head, n));
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 implementation to find the product of last
# 'n' nodes of the Linked List
 
''' 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
 
def reverseList(head_ref):
    next = None
    current = head_ref;
    prev = None;
 
    while (current != None):
        next = current.next;
        current.next = prev;
        prev = current;
        current = next;
    head_ref = prev
    return head_ref
 
# utility function to find the product of last 'n' nodes
def productOfLastN_NodesUtil(head, n):
 
    # if n == 0
    if (n <= 0):
        return 0;
 
    # reverse the linked list
    head = reverseList(head);
    prod = 1;
    current = head;
 
    # traverse the 1st 'n' nodes of the reversed
    # linked list and product them
    while (current != None and n):
         
        n -= 1
 
        # accumulate node's data to 'sum'
        prod *= current.data;
         
        # move to next node
        current = current.next;
         
    # reverse back the linked list
    head = reverseList(head);
 
    # required product
    return prod;
 
# Driver program to test above
if __name__=='__main__':
     
    head = None;
 
    # create 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);
 
    n = 2;
    print(productOfLastN_NodesUtil(head, n))
     
# This code is contributed by rutvik_56

C＃

// C# implementation to find
// the product of last 'n' 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;
}
 
static Node reverseList(Node head_ref)
{
    Node current, prev, next;
    current = head_ref;
    prev = null;
 
    while (current != null)
    {
        next = current.next;
        current.next = prev;
        prev = current;
        current = next;
    }
    head_ref = prev;
    return head_ref;
}
 
// utility function to find
// the product of last 'n' nodes
static int productOfLastN_NodesUtil(Node head, int n)
{
    // if n == 0
    if (n <= 0)
        return 0;
 
    // reverse the linked list
    head = reverseList(head);
 
    int prod = 1;
    Node current = head;
 
    // traverse the 1st 'n' nodes of the reversed
    // linked list and product them
    while (current != null && n-- >0)
    {
 
        // accumulate node's data to 'sum'
        prod *= current.data;
 
        // move to next node
        current = current.next;
    }
 
    // reverse back the linked list
    head = reverseList(head);
 
    // required product
    return prod;
}
 
// Driver Code
public static void Main(String[] args)
{
    Node head = null;
 
    // create 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 n = 2;
    Console.Write(productOfLastN_NodesUtil(head, n));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

// javascript implementation to find the product of last
// 'n' nodes of the Linked List
    /* 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 point to the new node */
        head_ref = new_node;
        return head_ref;
    }
 
    function reverseList(head_ref) {
var current, prev, next;
        current = head_ref;
        prev = null;
 
        while (current != null) {
            next = current.next;
            current.next = prev;
            prev = current;
            current = next;
        }
 
        head_ref = prev;
        return head_ref;
    }
 
    // utility function to find the product of last 'n' nodes
    function productOfLastN_NodesUtil(head , n) {
        // if n == 0
        if (n <= 0)
            return 0;
 
        // reverse the linked list
        head = reverseList(head);
 
        var prod = 1;
var current = head;
 
        // traverse the 1st 'n' nodes of the reversed
        // linked list and product them
        while (current != null && n-- > 0) {
 
            // accumulate node's data to 'sum'
            prod *= current.data;
 
            // move to next node
            current = current.next;
        }
 
        // reverse back the linked list
        head = reverseList(head);
 
        // required product
        return prod;
    }
 
    // Driver program to test above
     
var head = null;
 
        // create 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);
 
        var n = 2;
        document.write(productOfLastN_NodesUtil(head, n));
// This code contributed by aashish1995

输出：

48

时间复杂度： O(n)

方法4（使用链表的长度） ：

以下是步骤：

1. 计算给定链表的长度。让它成为len。
2. 首先从头开始遍历（len - n）个节点。
3. 然后遍历剩余的n个节点，同时遍历product它们。
4. 退回产品。

下面是上述方法的实现：

C++

// C++ implementation to find the product of last
// 'n' nodes of the Linked List
#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 = 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;
}
 
// utility function to find the product of last 'n' nodes
int productOfLastN_NodesUtil(struct Node* head, int n)
{
    // if n == 0
    if (n <= 0)
        return 0;
 
    int prod = 1, len = 0;
    struct Node* temp = head;
 
    // calculate the length of the linked list
    while (temp != NULL) {
        len++;
        temp = temp->next;
    }
 
    // count of first (len - n) nodes
    int c = len - n;
    temp = head;
 
    // just traverse the 1st 'c' nodes
    while (temp != NULL && c--)
 
        // move to next node
        temp = temp->next;
 
    // now traverse the last 'n' nodes and add them
    while (temp != NULL) {
 
        // accumulate node's data to sum
        prod *= temp->data;
 
        // move to next node
        temp = temp->next;
    }
 
    // required product
    return prod;
}
 
// Driver program to test above
int main()
{
    struct Node* head = NULL;
 
    // create linked list 10->6->8->4->12
    push(&head, 12);
    push(&head, 4);
    push(&head, 8);
    push(&head, 6);
    push(&head, 10);
 
    int n = 2;
    cout << productOfLastN_NodesUtil(head, n);
    return 0;
}

Java

// Java implementation to find the product of last
// 'n' nodes of the Linked List
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;
}
 
// utility function to find the product of last 'n' nodes
static int productOfLastN_NodesUtil(Node head, int n)
{
    // if n == 0
    if (n <= 0)
        return 0;
 
    int prod = 1, len = 0;
    Node temp = head;
 
    // calculate the length of the linked list
    while (temp != null)
    {
        len++;
        temp = temp.next;
    }
 
    // count of first (len - n) nodes
    int c = len - n;
    temp = head;
 
    // just traverse the 1st 'c' nodes
    while (temp != null && c-- >0)
 
        // move to next node
        temp = temp.next;
 
    // now traverse the last 'n' nodes and add them
    while (temp != null)
    {
        // accumulate node's data to sum
        prod *= temp.data;
 
        // move to next node
        temp = temp.next;
    }
 
    // required product
    return prod;
}
 
// Driver program to test above
public static void main(String[] args)
{
    Node head = null;
 
    // create 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 n = 2;
    System.out.print(productOfLastN_NodesUtil(head, n));
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation to find the product of last
# 'n' nodes of the Linked List
 
''' A Linked list node '''
class Node:
     
    def __init__(self):
        self.data = 0
        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();
 
    ''' 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;
 
# utility function to find the product of last 'n' nodes
def productOfLastN_NodesUtil(head, n):
 
    # if n == 0
    if (n <= 0):
        return 0;
 
    prod = 1
    len = 0;
    temp = head;
 
    # calculate the length of the linked list
    while (temp != None):   
        len += 1
        temp = temp.next;
     
    # count of first (len - n) nodes
    c = len - n;
    temp = head;
 
    # just traverse the 1st 'c' nodes
    while (temp != None and c > 0):
        c -= 1
 
        # move to next node
        temp = temp.next;
 
    # now traverse the last 'n' nodes and add them
    while (temp != None):
     
        # accumulate node's data to sum
        prod *= temp.data;
 
        # move to next node
        temp = temp.next;
    
    # required product
    return prod;
 
# Driver program to test above
if __name__=='__main__':  
    head = None;
 
    # create 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);
 
    n = 2;
    print(productOfLastN_NodesUtil(head, n));
 
# This code is contributed by Pratham76

C＃

// C# implementation to find the product of last
// 'n' nodes of the Linked List
using System;
 
public class GFG
{
  
/* A Linked list node */
public 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;
}
  
// utility function to find the product of last 'n' nodes
static int productOfLastN_NodesUtil(Node head, int n)
{
    // if n == 0
    if (n <= 0)
        return 0;
  
    int prod = 1, len = 0;
    Node temp = head;
  
    // calculate the length of the linked list
    while (temp != null)
    {
        len++;
        temp = temp.next;
    }
  
    // count of first (len - n) nodes
    int c = len - n;
    temp = head;
  
    // just traverse the 1st 'c' nodes
    while (temp != null && c-- >0)
  
        // move to next node
        temp = temp.next;
  
    // now traverse the last 'n' nodes and add them
    while (temp != null)
    {
        // accumulate node's data to sum
        prod *= temp.data;
  
        // move to next node
        temp = temp.next;
    }
  
    // required product
    return prod;
}
  
// Driver program to test above
public static void Main(String[] args)
{
    Node head = null;
  
    // create 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 n = 2;
    Console.Write(productOfLastN_NodesUtil(head, n));
}
}
 
// This code contributed by Rajput-Ji

Javascript

 
// Javascript implementation to find the product of last
// 'n' nodes of the Linked List
 
/* A Linked list node */
class Node
{
    constructor()
    {
        this.data = 0;
        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 point to the new node */
    head_ref = new_node;
    return head_ref;
}
  
// utility function to find the product of last 'n' nodes
function productOfLastN_NodesUtil(head, n)
{
    // if n == 0
    if (n <= 0)
        return 0;
  
    var prod = 1, len = 0;
    var temp = head;
  
    // calculate the length of the linked list
    while (temp != null)
    {
        len++;
        temp = temp.next;
    }
  
    // count of first (len - n) nodes
    var c = len - n;
    temp = head;
  
    // just traverse the 1st 'c' nodes
    while (temp != null && c-- >0)
  
        // move to next node
        temp = temp.next;
  
    // now traverse the last 'n' nodes and add them
    while (temp != null)
    {
        // accumulate node's data to sum
        prod *= temp.data;
  
        // move to next node
        temp = temp.next;
    }
  
    // required product
    return prod;
}
  
// Driver program to test above
var head = null;
 
// create 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);
 
var n = 2;
document.write(productOfLastN_NodesUtil(head, n));
 
 

输出：

48

时间复杂度： O(n)