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


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