Get Node Value

Problem Statement :

```This challenge is part of a tutorial track by MyCodeSchool

Given a pointer to the head of a linked list and a specific position, determine the data value at that position. Count backwards from the tail node. The tail is at postion 0, its parent is at 1 and so on.

Example
head refers to  3 -> 2 -> 1 -> 0 -> NULL
positionFromTail = 2

Each of the data values matches its distance from the tail. The value 2 is at the desired position.

Complete the getNode function in the editor below.

getNode has the following parameters:

int positionFromTail: the item to retrieve
Returns

int: the value at the desired position
Input Format

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

Each test case has the following format:
The first line contains an integer n, the number of elements in the linked list.
The next n lines contains an integer, the data value for an element of the linked list.
The last line contains an integer positionFromTail, the position from the tail to retrieve the value of.

Function Description```

Solution :

```                            ```Solution in C :

In C++ :

/*
Get Nth element from the end in a linked list of integers
Number of elements in the list will always be greater than N.
Node is defined as
struct Node
{
int data;
struct Node *next;
}
*/
{
// This is a "method-only" submission.
// You only need to complete this method.

int count = 0;

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

count = count-positionFromTail;

--count;

while(count!=0)
{
--count;
ptr=ptr->next;
}

return ptr->data;
}

In python3 :

nums = []

# First find tail of list as length is unknown
while current != None:
nums.append(current.data)
current = current.next

# Return value of item at given position
return nums[ len( nums ) - position - 1 ]

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;
}
*/

// This is a "method-only" submission.
// You only need to complete this method.

int count=0;
while(temp!=null)
{
count++;
temp=temp.next;
}

int k=count-n-1;
while(k>0)
{

k--;
temp=temp.next;
}
return temp.data;

}

In C :

// Complete the getNode function below.

/*
*
*     int data;
* };
*
*/
int c = 1, i = 1;
while(ptr->next != NULL){
c += 1;
ptr = ptr->next;
}
if(positionFromTail == 0)
return ptr->data;
c = c - positionFromTail;
while(i != c){
ptr = ptr->next;
i += 1;
}
return ptr->data;
}```
```

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