Find Merge Point of Two Lists


Problem Statement :


This challenge is part of a tutorial track by MyCodeSchool

Given pointers to the head nodes of 2 linked lists that merge together at some point, find the node where the two lists merge. The merge point is where both lists point to the same node, i.e. they reference the same memory location. It is guaranteed that the two head nodes will be different, and neither will be NULL. If the lists share a common node, return that node's data  value.

Note: After the merge point, both lists will share the same node pointers.

Example

In the diagram below, the two lists converge at Node x:

[List #1] a--->b--->c
                     \
                      x--->y--->z--->NULL
                     /
     [List #2] p--->q
Function Description

Complete the findMergeNode function in the editor below.

findMergeNode has the following parameters:

SinglyLinkedListNode pointer head1: a reference to the head of the first list
SinglyLinkedListNode pointer head2: a reference to the head of the second list

Returns

int: the data value of the node where the lists merge


Input Format

Do not read any input from stdin/console.

The first line contains an integer t, the number of test cases.


Solution :



title-img 


                            Solution in C :

In C :


// Complete the findMergeNode function below.

/*
 * For your reference:
 *
 * SinglyLinkedListNode {
 *     int data;
 *     SinglyLinkedListNode* next;
 * };
 *
 */
#include <math.h>
int findMergeNode(SinglyLinkedListNode *headA, SinglyLinkedListNode *headB)
{
    // Complete this function
    // Do not write the main method. 
    int countA = 0;
    int countB = 0;
    struct SinglyLinkedListNode *tempHeadA, *tempHeadB;
    tempHeadA = headA;
    tempHeadB = headB;
    
    while(tempHeadA != NULL){
        countA++;
        tempHeadA = tempHeadA->next;
    }
    while(tempHeadB != NULL){
        countB++;
        tempHeadB = tempHeadB->next;
    }
    
    // printf("%d %d", countA,countB);
    int biggerNodeExtraNodes = abs(countA - countB);
    // printf("%d", biggerNodeExtraNodes);
    while(biggerNodeExtraNodes--){
         printf("33333333333");
        if(countA > countB){
            headA = headA->next;
        } else{
             headB = headB->next;
        }
    }
    while(headA != NULL){
        printf("asdfghjklpoiuytredxdcfgh");
        if(headA == headB){
            return headA->data;
        }else{
             headA = headA->next;
             headB = headB->next;
        }
    }
    return 1;
}
                        


                        Solution in C++ :

In C++ :

/*
   Find merge point of two linked lists
   Node is defined as
   struct Node
   {
       int data;
       Node* next;
   }
*/
int getCount(Node* head)
{
  Node* current = head;
  int count = 0;
 
  while (current != NULL)
  {
    count++;
    current = current->next;
  }
 
  return count;
}

int getNode(int d, Node* head1, Node* head2)
{
  int i;
  Node* current1 = head1;
  Node* current2 = head2;
 
  for(i = 0; i < d; i++)
  {
    if(current1 == NULL)
    {  return -1; }
    current1 = current1->next;
  }
 
  while(current1 !=  NULL && current2 != NULL)
  {
    if(current1 == current2)
      return current1->data;
    current1= current1->next;
    current2= current2->next;
  }
 
  return -1;
}

int FindMergeNode(Node *headA, Node *headB)
{
    // Complete this function
    // Do not write the main method. 
    int c1 = getCount(headA);
  int c2 = getCount(headB);
  int d;
 
  if(c1 > c2)
  {
    d = c1 - c2;
    return getNode(d, headA, headB);
  }
  else
  {
    d = c2 - c1;
    return getNode(d, headB, headA);
  }
}
                    


                        Solution in Java :

In Java : 



/*
  Insert Node at the end of a linked list 
  head pointer input could be NULL as well for empty list
  Node is defined as 
  class Node {
     int data;
     Node next;
  }
*/
int FindMergeNode(Node headA, Node headB) {
    // Complete this function
    // Do not write the main method. 
      
    int countA=0;
    int countB=0;
    
    Node tempA=headA;
    Node tempB=headB;

    while(tempA!=null)
        {
        countA++;
        tempA=tempA.next;
    }
    
    
    while(tempB!=null)
        {
        countB++;
        tempB=tempB.next;
    }
    int diff=0;
    if(countA>countB)
        diff=countA-countB;
    else
        diff=countB-countA;
    tempA=headA;
    tempB=headB;
    if(countA>countB)
        {
         while(diff >0)
         {tempA=tempA.next;
         diff--;}
        
    }
    else
        {while(diff >0)
        { tempB=tempB.next;
          diff--;}
        
        
        }
    
    while(tempA!=null && tempB!=null)
    {
        
        tempA=tempA.next;
        tempB=tempB.next;
        if(tempA==tempB)
            return tempA.data;
        
    }
    return 0;
    

}
                    


                        Solution in Python : 
                            
In python3 :



"""
 Find the node at which both lists merge and return the data of that node.
 head could be None as well for empty list
 Node is defined as
 
 class Node(object):
 
   def __init__(self, data=None, next_node=None):
       self.data = data
       self.next = next_node

 
"""

def FindMergeNode(a, b):
    h = {}
    while a != None:
        h[a.data] = a.data
        a = a.next
    while b != None:
        if b.data in h:
            return h[b.data]
        b = b.next
    return None


View More Similar Problems

Insert a node at a specific position in a linked list

Given the pointer to the head node of a linked list and an integer to insert at a certain position, create a new node with the given integer as its data attribute, insert this node at the desired position and return the head node. A position of 0 indicates head, a position of 1 indicates one node away from the head and so on. The head pointer given may be null meaning that the initial list is e

View Solution →

Delete a Node

Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value. Example: list=0->1->2->3 position=2 After removing the node at position 2, list'= 0->1->-3. Function Description: Complete the deleteNode function in the editor below. deleteNo

View Solution →

Print in Reverse

Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing

View Solution →

Reverse a linked list

Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio

View Solution →

Compare two linked lists

You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lis

View Solution →

Merge two sorted linked lists

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the heads of two sorted linked lists, merge them into a single, sorted linked list. Either head pointer may be null meaning that the corresponding list is empty. Example headA refers to 1 -> 3 -> 7 -> NULL headB refers to 1 -> 2 -> NULL The new list is 1 -> 1 -> 2 -> 3 -> 7 -> NULL. Function Description C

View Solution →