Linked list

Question

How does this code return Nth the node from the end of the linked list in one pass?

Node * GetNthNode ( Node* Head , int NthNode )

{

      Node * pNthNode = NULL;

      Node * pTempNode = NULL;

      int nCurrentElement = 0;

     

for ( pTempNode = Head; pTempNode != NULL; pTempNode = pTempNode->pNext )

      {

            nCurrentElement++;                 

            if ( nCurrentElement - NthNode == 0 )

            {

                  pNthNode = Head;

            }

            else

            if ( nCurrentElement - NthNode > 0)

            {

                  pNthNode = pNthNode ->pNext;

            }                            

      }

      if (pNthNode )

      {

            return pNthNode;

      }

      else

            return NULL;

}

Comments

format the code please :)

Answer 1

 Size of linked list is not given here. Below is the generic logic

Distance from head to K and A to END is same.

Step 1 in the image describes the first if condition :

if (nCurrentElement - NthNode == 0 ) {
    pNthNode = Head;
}
//pNthNode = NewPointer

Step 2 in the image describes elseif condition :

else if ( nCurrentElement - NthNode > 0){
    pNthNode = pNthNode ->pNext;
}

By the end of for loop NewPointer is the Nth node from end.