查找两个单链表中的公共节点

给定两个链表，任务是找出两个单链表中公共节点的数量。

例子：

Input: List A = 3 -> 4 -> 12 -> 10 -> 17, List B = 10 -> 4 -> 8 -> 575 -> 34 -> 12 
Output: Number of common nodes in both list is = 3

Input: List A = 12 -> 4 -> 65 -> 14 -> 59, List B = 14 -> 15 -> 23 -> 17 -> 41 -> 54 
Output: Number of common nodes in both list is = 1

朴素方法：将列表 A 的每个节点与列表 B 的每个节点进行比较。如果节点匹配，则增加计数并在比较所有节点后返回计数。

以下是上述方法的实现：

C++

// C++ implementation of above approach
#include 
using namespace std;
 
// Structure of a linked list node
struct Node {
    int data;
    struct Node* next;
};
 
// Function to common nodes which have
// same value node(s) both list
int countCommonNodes(struct Node* head1, struct Node* head2)
{
    // list A
    struct Node* current1 = head1;
 
    // list B
    struct Node* current2 = head2;
 
    // set count = 0
    int count = 0;
 
    // traverse list A till the end of list
    while (current1 != NULL) {
 
        // traverse list B till the end of list
        while (current2 != NULL) {
 
            // if data is match then count increase
            if (current1->data == current2->data)
                count++;
 
            // increase current pointer for next node
            current2 = current2->next;
        }
 
        // increase current pointer of list A
        current1 = current1->next;
 
        // initialize list B starting point
        current2 = head2;
    }
 
    // return count
    return count;
}
 
/* Utility function to insert a node at the beginning */
void push(struct Node** head_ref, int new_data)
{
    struct Node* new_node = new Node;
    new_node->data = new_data;
    new_node->next = *head_ref;
    *head_ref = new_node;
}
 
/* Utility function to print a linked list */
void printList(struct Node* head)
{
    while (head != NULL) {
        cout << head->data << " ";
        head = head->next;
    }
    cout << endl;
}
 
/* Driver program to test above functions */
int main()
{
    struct Node* head1 = NULL;
    struct Node* head2 = NULL;
 
    /* Create following linked list
    List A = 3 -> 4 -> 12 -> 10 -> 17
*/
 
    push(&head1, 17);
    push(&head1, 10);
    push(&head1, 12);
    push(&head1, 4);
    push(&head1, 3);
 
    // List B = 10 -> 4 -> 8 -> 575 -> 34 -> 12
    push(&head2, 12);
    push(&head2, 34);
    push(&head2, 575);
    push(&head2, 8);
    push(&head2, 4);
    push(&head2, 10);
 
    // print list A
    cout << "Given Linked List A: \n";
    printList(head1);
 
    // print list B
    cout << "Given Linked List B: \n";
    printList(head2);
 
    // call function for count common node
    int count = countCommonNodes(head1, head2);
 
    // print number of common node in both list
    cout << "Number of common node in both list is = " << count;
    return 0;
}

Java

// Java implementation of above approach
class GFG {
 
    // Structure of a linked list node
    static class Node {
        int data;
        Node next;
    };
 
    // Function to common nodes which have
    // same value node(s) both list
    static int countCommonNodes(Node head1, Node head2)
    {
        // list A
        Node current1 = head1;
 
        // list B
        Node current2 = head2;
 
        // set count = 0
        int count = 0;
 
        // traverse list A till the end of list
        while (current1 != null) {
 
            // traverse list B till the end of list
            while (current2 != null) {
 
                // if data is match then count increase
                if (current1.data == current2.data)
                    count++;
 
                // increase current pointer for next node
                current2 = current2.next;
            }
 
            // increase current pointer of list A
            current1 = current1.next;
 
            // initialize list B starting point
            current2 = head2;
        }
 
        // return count
        return count;
    }
 
    // Utility function to insert a node at the beginning /
    static Node push(Node head_ref, int new_data)
    {
        Node new_node = new Node();
        new_node.data = new_data;
        new_node.next = head_ref;
        head_ref = new_node;
        return head_ref;
    }
 
    // Utility function to print a linked list
    static void printList(Node head)
    {
        while (head != null) {
            System.out.print(head.data + " ");
            head = head.next;
        }
        System.out.println();
    }
 
    // Driver code
    public static void main(String args[])
    {
        Node head1 = null;
        Node head2 = null;
 
        // Create following linked list
        // List A = 3 . 4 . 12 . 10 . 17
 
        head1 = push(head1, 17);
        head1 = push(head1, 10);
        head1 = push(head1, 12);
        head1 = push(head1, 4);
        head1 = push(head1, 3);
 
        // List B = 10 . 4 . 8 . 575 . 34 . 12
        head2 = push(head2, 12);
        head2 = push(head2, 34);
        head2 = push(head2, 575);
        head2 = push(head2, 8);
        head2 = push(head2, 4);
        head2 = push(head2, 10);
 
        // print list A
        System.out.print("Given Linked List A: \n");
        printList(head1);
 
        // print list B
        System.out.print("Given Linked List B: \n");
        printList(head2);
 
        // call function for count common node
        int count = countCommonNodes(head1, head2);
 
        // print number of common node in both list
        System.out.print("Number of common node in both list is = " + count);
    }
}
 
// This code is contributed by Arnab Kundu

Python3

# Python3 implementation of above approach
 
# Link list node
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None
 
# Function to common nodes which have
# same value node(s) both list
def countCommonNodes(head1, head2):
     
    # list A
    current1 = head1
 
    # list B
    current2 = head2
 
    # set count = 0
    count = 0
 
    # traverse list A till the end of list
    while (current1 != None):
 
        # traverse list B till the end of list
        while (current2 != None):
 
            # if data is match then count increase
            if (current1.data == current2.data):
                count = count + 1
 
            # increase current pointer for next node
            current2 = current2.next
         
        # increase current pointer of list A
        current1 = current1.next
 
        # initialize list B starting point
        current2 = head2
     
    # return count
    return count
 
# Utility function to insert a node at the beginning
def push(head_ref, new_data):
    new_node = Node(0)
    new_node.data = new_data
    new_node.next = head_ref
    head_ref = new_node
    return head_ref
 
# Utility function to print a linked list
def printList( head):
    while (head != None):
        print(head.data, end = " ")
        head = head.next
     
    print("")
 
# Driver Code
if __name__=='__main__':
 
    head1 = None
    head2 = None
 
    # Create following linked list
    # List A = 3 . 4 . 12 . 10 . 17
    head1 = push(head1, 17)
    head1 = push(head1, 10)
    head1 = push(head1, 12)
    head1 = push(head1, 4)
    head1 = push(head1, 3)
 
    # List B = 10 . 4 . 8 . 575 . 34 . 12
    head2 = push(head2, 12)
    head2 = push(head2, 34)
    head2 = push(head2, 575)
    head2 = push(head2, 8)
    head2 = push(head2, 4)
    head2 = push(head2, 10)
 
    # print list A
    print("Given Linked List A: ")
    printList(head1)
 
    # print list B
    print("Given Linked List B: ")
    printList(head2)
 
    # call function for count common node
    count = countCommonNodes(head1, head2)
 
    # print number of common node in both list
    print("Number of common node in both list is = ", count)
 
# This code is contributed by Arnab Kundu

C#

// C# implementation of the approach
using System;
 
class GFG {
 
    // Structure of a linked list node
    public class Node {
        public int data;
        public Node next;
    };
 
    // Function to common nodes which have
    // same value node(s) both list
    static int countCommonNodes(Node head1, Node head2)
    {
        // list A
        Node current1 = head1;
 
        // list B
        Node current2 = head2;
 
        // set count = 0
        int count = 0;
 
        // traverse list A till the end of list
        while (current1 != null) {
 
            // traverse list B till the end of list
            while (current2 != null) {
 
                // if data is match then count increase
                if (current1.data == current2.data)
                    count++;
 
                // increase current pointer for next node
                current2 = current2.next;
            }
 
            // increase current pointer of list A
            current1 = current1.next;
 
            // initialize list B starting point
            current2 = head2;
        }
 
        // return count
        return count;
    }
 
    // Utility function to insert a node at the beginning /
    static Node push(Node head_ref, int new_data)
    {
        Node new_node = new Node();
        new_node.data = new_data;
        new_node.next = head_ref;
        head_ref = new_node;
        return head_ref;
    }
 
    // Utility function to print a linked list
    static void printList(Node head)
    {
        while (head != null) {
            Console.Write(head.data + " ");
            head = head.next;
        }
        Console.WriteLine();
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        Node head1 = null;
        Node head2 = null;
 
        // Create following linked list
        // List A = 3 . 4 . 12 . 10 . 17
 
        head1 = push(head1, 17);
        head1 = push(head1, 10);
        head1 = push(head1, 12);
        head1 = push(head1, 4);
        head1 = push(head1, 3);
 
        // List B = 10 . 4 . 8 . 575 . 34 . 12
        head2 = push(head2, 12);
        head2 = push(head2, 34);
        head2 = push(head2, 575);
        head2 = push(head2, 8);
        head2 = push(head2, 4);
        head2 = push(head2, 10);
 
        // print list A
        Console.Write("Given Linked List A: \n");
        printList(head1);
 
        // print list B
        Console.Write("Given Linked List B: \n");
        printList(head2);
 
        // call function for count common node
        int count = countCommonNodes(head1, head2);
 
        // print number of common node in both list
        Console.Write("Number of common node in both list is = " + count);
    }
}
 
/* This code contributed by PrinciRaj1992 */

Javascript

 
// Javascript implementation of above approach   
// Structure of a linked list node
class Node
{
    constructor(val)
    {
        this.data = val;
        this.next = null;
    }
}
 
// Function to common nodes which have
// same value node(s) both list
function countCommonNodes(head1, head2)
{
     
    // List A
    var current1 = head1;
 
    // List B
    var current2 = head2;
 
    // Set count = 0
    var count = 0;
 
    // Traverse list A till the end of list
    while (current1 != null)
    {
         
        // Traverse list B till the end of list
        while (current2 != null)
        {
             
            // If data is match then count increase
            if (current1.data == current2.data)
                count++;
 
            // Increase current pointer for next node
            current2 = current2.next;
        }
 
        // Increase current pointer of list A
        current1 = current1.next;
 
        // Initialize list B starting point
        current2 = head2;
    }
 
    // Return count
    return count;
}
 
// Utility function to insert a node at the beginning
function push(head_ref, new_data)
{
    var new_node = new Node();
    new_node.data = new_data;
    new_node.next = head_ref;
    head_ref = new_node;
    return head_ref;
}
 
// Utility function to print a linked list
function printList(head)
{
    while (head != null)
    {
        document.write(head.data + " ");
        head = head.next;
    }
    document.write("<br/>");
}
 
// Driver code
var head1 = null;
var head2 = null;
 
// Create following linked list
// List A = 3 . 4 . 12 . 10 . 17
head1 = push(head1, 17);
head1 = push(head1, 10);
head1 = push(head1, 12);
head1 = push(head1, 4);
head1 = push(head1, 3);
 
// List B = 10 . 4 . 8 . 575 . 34 . 12
head2 = push(head2, 12);
head2 = push(head2, 34);
head2 = push(head2, 575);
head2 = push(head2, 8);
head2 = push(head2, 4);
head2 = push(head2, 10);
 
// Print list A
document.write("Given Linked List A: <br/>");
printList(head1);
 
// Print list B
document.write("Given Linked List B: <br/>");
printList(head2);
 
// Call function for count common node
var count = countCommonNodes(head1, head2);
 
// Print number of common node in both list
document.write("Number of common node in both list is = " +
               count);
 
// This code is contributed by gauravrajput1
 

输出：

Given Linked List A: 
3 4 12 10 17 
Given Linked List B: 
10 4 8 575 34 12 
Number of common node in both list is = 3

时间复杂度： O(M*N)，其中 M 是列表 A 的长度，N 是列表 B 的长度
高效的解决方案：将链表A的所有节点插入unordered_set，然后在unordered_set中检查链表B的每个节点。如果找到增加计数并在最后返回计数。

下面是上述方法的实现：

C++

// C++ implementation of above approach
#include 
using namespace std;
 
// Structure of a linked list node
struct Node {
    int data;
    struct Node* next;
};
 
// Function to common nodes which have
// same value node(s) both list
int countCommonNodes(struct Node* head1, struct Node* head2)
{
    // list  A
    struct Node* current1 = head1;
 
    // list B
    struct Node* current2 = head2;
 
    // set count  = 0
    int count = 0;
 
    // create unordered_set
    unordered_set map;
 
    // traverse list A till the end of list
    while (current1 != NULL) {
 
        // insert list data in map
        map.insert(current1->data);
 
        // increase current pointer of list A
        current1 = current1->next;
    }
 
    while (current2 != NULL) {
 
        // traverse list B till the end of list
        // if data match then increase count
        if (map.find(current2->data) != map.end())
            count++;
 
        // increase current pointer of list B
        current2 = current2->next;
    }
 
    // return count
    return count;
}
 
/* Utility function to insert a node at the beginning */
void push(struct Node** head_ref, int new_data)
{
    struct Node* new_node = new Node;
    new_node->data = new_data;
    new_node->next = *head_ref;
    *head_ref = new_node;
}
 
/* Utility function to print a linked list */
void printList(struct Node* head)
{
    while (head != NULL) {
        cout << head->data << " ";
        head = head->next;
    }
    cout << endl;
}
 
/* Driver program to test above functions */
int main()
{
    struct Node* head1 = NULL;
    struct Node* head2 = NULL;
 
    /* Create following linked list
    List A = 3 -> 4 -> 12 -> 10 -> 17
   */
 
    push(&head1, 17);
    push(&head1, 10);
    push(&head1, 12);
    push(&head1, 4);
    push(&head1, 3);
 
    // List B = 10 -> 4 -> 8 -> 575 -> 34 -> 12
    push(&head2, 12);
    push(&head2, 34);
    push(&head2, 575);
    push(&head2, 8);
    push(&head2, 4);
    push(&head2, 10);
 
    // print list A
    cout << "Given Linked List A: \n";
    printList(head1);
 
    // print list B
    cout << "Given Linked List B: \n";
    printList(head2);
 
    // call function for count common node
    int count = countCommonNodes(head1, head2);
 
    // print number of common node in both list
    cout << "Number of common node in both list is = " << count;
 
    return 0;
}

Java

// Java implementation of above approach
import java.util.*;
class solution {
 
    // static class of a linked list node
    static class Node {
        int data;
        Node next;
    }
 
    // Function to common nodes which have
    // same value node(s) both list
    static int countCommonNodes(Node head1, Node head2)
    {
        // list  A
        Node current1 = head1;
 
        // list B
        Node current2 = head2;
 
        // set count  = 0
        int count = 0;
 
        // create vector
        Vector map = new Vector();
 
        // traverse list A till the end of list
        while (current1 != null) {
 
            // insert list data in map
            map.add(current1.data);
 
            // increase current pointer of list A
            current1 = current1.next;
        }
 
        while (current2 != null) {
 
            // traverse list B till the end of list
            // if data match then increase count
            if (map.contains(current2.data))
                count++;
 
            // increase current pointer of list B
            current2 = current2.next;
        }
 
        // return count
        return count;
    }
 
    /* Utility function to insert a node at the beginning */
    static Node push(Node head_ref, int new_data)
    {
        Node new_node = new Node();
        new_node.data = new_data;
        new_node.next = head_ref;
        head_ref = new_node;
 
        return head_ref;
    }
 
    /* Utility function to print a linked list */
    static void printList(Node head)
    {
        while (head != null) {
            System.out.print(head.data + " ");
            head = head.next;
        }
        System.out.println();
    }
 
    /* Driver program to test above functions */
    public static void main(String args[])
    {
        Node head1 = null;
        Node head2 = null;
 
        /* Create following linked list
    List A = 3 . 4 . 12 . 10 . 17
   */
 
        head1 = push(head1, 17);
        head1 = push(head1, 10);
        head1 = push(head1, 12);
        head1 = push(head1, 4);
        head1 = push(head1, 3);
 
        // List B = 10 . 4 . 8 . 575 . 34 . 12
        head2 = push(head2, 12);
        head2 = push(head2, 34);
        head2 = push(head2, 575);
        head2 = push(head2, 8);
        head2 = push(head2, 4);
        head2 = push(head2, 10);
 
        // print list A
        System.out.print("Given Linked List A: \n");
        printList(head1);
 
        // print list B
        System.out.print("Given Linked List B: \n");
        printList(head2);
 
        // call function for count common node
        int count = countCommonNodes(head1, head2);
 
        // print number of common node in both list
        System.out.print("Number of common node in both list is = " + count);
    }
}
// This code is contributed by Arnab Kundu

Python3

# Python3 implementation of the approach
 
# Structure of a linked list node
class Node:
    def __init__(self):
        self.data = 0
        self.next = None
 
# Function to common nodes which have
# same value node(s) both list
def countCommonNodes(head1, head2):
 
    # List A
    current1 = head1
 
    # List B
    current2 = head2
 
    # Set count = 0
    count = 0
 
    # Create unordered_set
    map_=set()
 
    # Traverse list A till the end of list
    while (current1 != None) :
 
        # Add list data in map_
        map_.add(current1.data)
 
        # Increase current pointer of list A
        current1 = current1.next
     
    while (current2 != None) :
 
        # Traverse list B till the end of list
        # if data match then increase count
        if ((current2.data) in map_):
            count = count + 1
 
        # Increase current pointer of list B
        current2 = current2.next
     
    # Return count
    return count
 
# Utility function to add a node at the beginning
def push(head_ref,new_data):
    new_node = Node()
    new_node.data = new_data
    new_node.next = head_ref
    head_ref = new_node
    return head_ref
 
# Utility function to print a linked list
def printList(head):
    while (head != None) :
        print(head.data, end = " ")
        head = head.next
     
# Driver program to test above functions
head1 = None
head2 = None
 
# Create following linked list
# List A = 3 . 4 . 12 . 10 . 17
head1 = push(head1, 17)
head1 = push(head1, 10)
head1 = push(head1, 12)
head1 = push(head1, 4)
head1 = push(head1, 3)
 
# List B = 10 . 4 . 8 . 575 . 34 . 12
head2 = push(head2, 12)
head2 = push(head2, 34)
head2 = push(head2, 575)
head2 = push(head2, 8)
head2 = push(head2, 4)
head2 = push(head2, 10)
 
# Print list A
print("Given Linked List A: ")
printList(head1)
 
# Print list B
print( "\nGiven Linked List B: ")
printList(head2)
 
# Call function for count common node
count = countCommonNodes(head1, head2)
 
# Print number of common node in both list
print("\nNumber of common node in both list is = " , count)
 
# This code is contributed by Arnab Kundu

C＃

// C# implementation of above approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
    // static class of a linked list node
    public class Node
    {
        public int data;
        public Node next;
    }
 
    // Function to common nodes which have
    // same value node(s) both list
    static int countCommonNodes(Node head1, Node head2)
    {
        // list A
        Node current1 = head1;
 
        // list B
        Node current2 = head2;
 
        // set count = 0
        int count = 0;
 
        // create vector
        List map = new List();
 
        // traverse list A till the end of list
        while (current1 != null)
        {
 
            // insert list data in map
            map.Add(current1.data);
 
            // increase current pointer of list A
            current1 = current1.next;
        }
 
        while (current2 != null)
        {
 
            // traverse list B till the end of list
            // if data match then increase count
            if (map.Contains(current2.data))
                count++;
 
            // increase current pointer of list B
            current2 = current2.next;
        }
 
        // return count
        return count;
    }
 
    /* Utility function to insert a node at the beginning */
    static Node push(Node head_ref, int new_data)
    {
        Node new_node = new Node();
        new_node.data = new_data;
        new_node.next = head_ref;
        head_ref = new_node;
 
        return head_ref;
    }
 
    /* Utility function to print a linked list */
    static void printList(Node head)
    {
        while (head != null)
        {
            Console.Write(head.data + " ");
            head = head.next;
        }
        Console.WriteLine();
    }
 
    /* Driver code */
    public static void Main(String []args)
    {
        Node head1 = null;
        Node head2 = null;
 
        /* Create following linked list
    List A = 3 . 4 . 12 . 10 . 17
*/
 
        head1 = push(head1, 17);
        head1 = push(head1, 10);
        head1 = push(head1, 12);
        head1 = push(head1, 4);
        head1 = push(head1, 3);
 
        // List B = 10 . 4 . 8 . 575 . 34 . 12
        head2 = push(head2, 12);
        head2 = push(head2, 34);
        head2 = push(head2, 575);
        head2 = push(head2, 8);
        head2 = push(head2, 4);
        head2 = push(head2, 10);
 
        // print list A
        Console.Write("Given Linked List A: \n");
        printList(head1);
 
        // print list B
        Console.Write("Given Linked List B: \n");
        printList(head2);
 
        // call function for count common node
        int count = countCommonNodes(head1, head2);
 
        // print number of common node in both list
        Console.Write("Number of common node in both list is = " + count);
    }
}
 
// This code has been contributed by 29AjayKumar

Javascript

      // JavaScript implementation of above approach
      // static class of a linked list node
      class Node {
        constructor() {
          this.data = 0;
          this.next = null;
        }
      }
 
      // Function to common nodes which have
      // same value node(s) both list
      function countCommonNodes(head1, head2) {
        // list A
        var current1 = head1;
 
        // list B
        var current2 = head2;
 
        // set count = 0
        var count = 0;
 
        // create vector
        var map = [];
 
        // traverse list A till the end of list
        while (current1 != null) {
          // insert list data in map
          map.push(current1.data);
 
          // increase current pointer of list A
          current1 = current1.next;
        }
 
        while (current2 != null) {
          // traverse list B till the end of list
          // if data match then increase count
          if (map.includes(current2.data)) count++;
 
          // increase current pointer of list B
          current2 = current2.next;
        }
 
        // return count
        return count;
      }
 
      /* Utility function to insert a node at the beginning */
      function push(head_ref, new_data) {
        var new_node = new Node();
        new_node.data = new_data;
        new_node.next = head_ref;
        head_ref = new_node;
 
        return head_ref;
      }
 
      /* Utility function to print a linked list */
      function printList(head) {
        while (head != null) {
          document.write(head.data + " ");
          head = head.next;
        }
        document.write("<br>");
      }
 
      /* Driver code */
      var head1 = null;
      var head2 = null;
 
      /* Create following linked list
    List A = 3 . 4 . 12 . 10 . 17
*/
 
      head1 = push(head1, 17);
      head1 = push(head1, 10);
      head1 = push(head1, 12);
      head1 = push(head1, 4);
      head1 = push(head1, 3);
 
      // List B = 10 . 4 . 8 . 575 . 34 . 12
      head2 = push(head2, 12);
      head2 = push(head2, 34);
      head2 = push(head2, 575);
      head2 = push(head2, 8);
      head2 = push(head2, 4);
      head2 = push(head2, 10);
 
      // print list A
      document.write("Given Linked List A: <br>");
      printList(head1);
 
      // print list B
      document.write("Given Linked List B: <br>");
      printList(head2);
 
      // call function for count common node
      var count = countCommonNodes(head1, head2);
 
      // print number of common node in both list
      document.write("Number of common node in both list is = " + count);
       
      // This code is contributed by rdtank.
    

输出：

Given Linked List A: 
3 4 12 10 17 
Given Linked List B: 
10 4 8 575 34 12 
Number of common node in both list is = 3

时间复杂度： O(N)
空间复杂度： O(N)