# 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:

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 :

```                            ```Solution in C :

In C++ :

/*
Find merge point of two linked lists
Node is defined as
struct Node
{
int data;
Node* next;
}
*/
{
int count = 0;

while (current != NULL)
{
count++;
current = current->next;
}

return count;
}

{
int i;

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;
}

{
// Complete this function
// Do not write the main method.
int d;

if(c1 > c2)
{
d = c1 - c2;
}
else
{
d = c2 - c1;
}
}

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;
}
*/
// Complete this function
// Do not write the main method.

int countA=0;
int countB=0;

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;
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;

}

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

In C :

// Complete the findMergeNode function below.

/*
*
*     int data;
* };
*
*/
#include <math.h>
{
// Complete this function
// Do not write the main method.
int countA = 0;
int countB = 0;

countA++;
}
countB++;
}

// printf("%d %d", countA,countB);
int biggerNodeExtraNodes = abs(countA - countB);
// printf("%d", biggerNodeExtraNodes);
while(biggerNodeExtraNodes--){
printf("33333333333");
if(countA > countB){
} else{
}
}
printf("asdfghjklpoiuytredxdcfgh");
}else{
}
}
return 1;
}```
```

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