# 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 :

// 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;
}```
```

```                        ```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;
}
}```
```

```                        ```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;
}
*/
// 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;

}```
```

```                        ```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```
```

## 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

## 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

## 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