在不求长度的情况下查找两个链表的交点

一个系统中有两个单链表。由于一些编程错误，其中一个链表的末端节点链接到了第二个链表，形成了一个倒 Y 形链表。编写一个程序来获得两个链表合并的点。
例子：

Input: 1 -> 2 -> 3 -> 4 -> 5 -> 6
                      ^
                      |
            7 -> 8 -> 9
Output: 4

Input:         13 -> 14 -> 5 -> 6
                      ^
                      |
      10 -> 2 -> 3 -> 4 
Output: 14

先决条件：编写一个函数来获取两个链表的交点
方法：取两个指针指向两个链表的头部。如果其中一个更早到达末尾，则通过将其移动到另一个列表的开头来使用它。一旦它们都经过重新分配，它们将与碰撞点等距。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
/* Link list node */
class Node {
public:
    int data;
    Node* next;
};
 
// Function to return the intersection point
// of the two linked lists head1 and head2
int getIntesectionNode(Node* head1, Node* head2)
{
    Node* current1 = head1;
    Node* current2 = head2;
 
    // If one of the head is NULL
    if (!current1 or !current2)
        return -1;
 
    // Continue until we find intersection node
    while (current1 and current2
           and current1 != current2) {
        current1 = current1->next;
        current2 = current2->next;
 
        // If we get intersection node
        if (current1 == current2)
            return current1->data;
 
        // If one of them reaches end
        if (!current1)
            current1 = head2;
        if (!current2)
            current2 = head1;
    }
 
    return current1->data;
}
 
// Driver code
int main()
{
    /*
        Create two linked lists
     
        1st 3->6->9->15->30
        2nd 10->15->30
     
        15 is the intersection point
    */
 
    Node* newNode;
 
    // Addition of new nodes
    Node* head1 = new Node();
    head1->data = 10;
 
    Node* head2 = new Node();
    head2->data = 3;
 
    newNode = new Node();
    newNode->data = 6;
    head2->next = newNode;
 
    newNode = new Node();
    newNode->data = 9;
    head2->next->next = newNode;
 
    newNode = new Node();
    newNode->data = 15;
    head1->next = newNode;
    head2->next->next->next = newNode;
 
    newNode = new Node();
    newNode->data = 30;
    head1->next->next = newNode;
 
    head1->next->next->next = NULL;
 
    cout << getIntesectionNode(head1, head2);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
 
/* Link list node */
static class Node
{
 
    int data;
    Node next;
};
 
// Function to return the intersection point
// of the two linked lists head1 and head2
static int getIntesectionNode(Node head1, Node head2)
{
    Node current1 = head1;
    Node current2 = head2;
 
    // If one of the head is null
    if (current1 == null || current2 == null )
        return -1;
 
    // Continue until we find intersection node
    while (current1 != null && current2 != null
        && current1 != current2)
    {
        current1 = current1.next;
        current2 = current2.next;
 
        // If we get intersection node
        if (current1 == current2)
            return current1.data;
 
        // If one of them reaches end
        if (current1 == null )
            current1 = head2;
        if (current2 == null )
            current2 = head1;
    }
 
    return current1.data;
}
 
// Driver code
public static void main(String[] args)
{
    /*
        Create two linked lists
     
        1st 3.6.9.15.30
        2nd 10.15.30
     
        15 is the intersection point
    */
 
    Node newNode;
 
    // Addition of new nodes
    Node head1 = new Node();
    head1.data = 10;
 
    Node head2 = new Node();
    head2.data = 3;
 
    newNode = new Node();
    newNode.data = 6;
    head2.next = newNode;
 
    newNode = new Node();
    newNode.data = 9;
    head2.next.next = newNode;
 
    newNode = new Node();
    newNode.data = 15;
    head1.next = newNode;
    head2.next.next.next = newNode;
 
    newNode = new Node();
    newNode.data = 30;
    head1.next.next = newNode;
 
    head1.next.next.next = null;
 
    System.out.print(getIntesectionNode(head1, head2));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
 
''' Link list node '''
class new_Node:
         
    # Constructor to initialize the node object
    def __init__(self, data):
        self.data = data
        self.next = None
 
# Function to return the intersection point
# of the two linked lists head1 and head2
def getIntesectionNode(head1, head2):
     
    current1 = head1
    current2 = head2
     
    # If one of the head is None
    if (not current1 or not current2 ):
        return -1
         
    # Continue until we find intersection node
    while (current1 and current2 and current1 != current2):
        current1 = current1.next
        current2 = current2.next
         
        # If we get intersection node
        if (current1 == current2):
            return current1.data
             
        # If one of them reaches end
        if (not current1):
            current1 = head2
         
        if (not current2):
            current2 = head1
             
    return current1.data
 
# Driver code
'''
    Create two linked lists
 
    1st 3.6.9.15.30
    2nd 10.15.30
 
    15 is the intersection po
'''
 
# Addition of newNodes
head1 = new_Node(10)
 
head2 = new_Node(3)
 
newNode = new_Node(6)
head2.next = newNode
 
newNode = new_Node(9)
head2.next.next = newNode
 
newNode = new_Node(15)
head1.next = newNode
head2.next.next.next = newNode
 
newNode = new_Node(30)
head1.next.next = newNode
 
head1.next.next.next = None
 
print(getIntesectionNode(head1, head2))
 
# This code is contributed by shubhamsingh10

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
/* Link list node */
class Node
{
 
    public int data;
    public Node next;
};
 
// Function to return the intersection point
// of the two linked lists head1 and head2
static int getIntesectionNode(Node head1, Node head2)
{
    Node current1 = head1;
    Node current2 = head2;
 
    // If one of the head is null
    if (current1 == null || current2 == null )
        return -1;
 
    // Continue until we find intersection node
    while (current1 != null && current2 != null
        && current1 != current2)
    {
        current1 = current1.next;
        current2 = current2.next;
 
        // If we get intersection node
        if (current1 == current2)
            return current1.data;
 
        // If one of them reaches end
        if (current1 == null )
            current1 = head2;
        if (current2 == null )
            current2 = head1;
    }
 
    return current1.data;
}
 
// Driver code
public static void Main(String[] args)
{
    /*
        Create two linked lists
     
        1st 3.6.9.15.30
        2nd 10.15.30
     
        15 is the intersection point
    */
 
    Node newNode;
 
    // Addition of new nodes
    Node head1 = new Node();
    head1.data = 10;
 
    Node head2 = new Node();
    head2.data = 3;
 
    newNode = new Node();
    newNode.data = 6;
    head2.next = newNode;
 
    newNode = new Node();
    newNode.data = 9;
    head2.next.next = newNode;
 
    newNode = new Node();
    newNode.data = 15;
    head1.next = newNode;
    head2.next.next.next = newNode;
 
    newNode = new Node();
    newNode.data = 30;
    head1.next.next = newNode;
 
    head1.next.next.next = null;
 
    Console.Write(getIntesectionNode(head1, head2));
}
}
 
// This code is contributed by PrinciRaj1992

Javascript

// Javascript implementation of the approach
 
/* Link list node */
class Node {
 
    constructor()
    {
        this.data = 0;
        this.next = null;
    }
};
 
// Function to return the intersection point
// of the two linked lists head1 and head2
function getIntesectionNode(head1, head2)
{
    var current1 = head1;
    var current2 = head2;
 
    // If one of the head is null
    if (!current1 || !current2)
        return -1;
 
    // Continue until we find intersection node
    while (current1 && current2
           && current1 != current2) {
        current1 = current1.next;
        current2 = current2.next;
 
        // If we get intersection node
        if (current1 == current2)
            return current1.data;
 
        // If one of them reaches end
        if (!current1)
            current1 = head2;
        if (!current2)
            current2 = head1;
    }
 
    return current1.data;
}
 
// Driver code
/*
    Create two linked lists
 
    1st 3.6.9.15.30
    2nd 10.15.30
 
    15 is the intersection point
*/
var newNode;
 
// Addition of new nodes
var head1 = new Node();
head1.data = 10;
var head2 = new Node();
head2.data = 3;
newNode = new Node();
newNode.data = 6;
head2.next = newNode;
newNode = new Node();
newNode.data = 9;
head2.next.next = newNode;
newNode = new Node();
newNode.data = 15;
head1.next = newNode;
head2.next.next.next = newNode;
newNode = new Node();
newNode.data = 30;
head1.next.next = newNode;
head1.next.next.next = null;
document.write( getIntesectionNode(head1, head2));
 
// This code is contributed by noob2000.

输出：

15

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。