Question

In C++

Assume entries in a linked list are of type struct listrec:

struct listrec

{

            struct listrec    *prev;

float                 value;

            struct listrec    *next;  

};

listrec *head, *tail;

NULL 1.0 3.0 value next value next prev prev Value next prev head tail NULL

Write a main() routine in which the user is asked the number of nodes to create in the list (number greater than or equal to zero) then create the following type of linked list (use a loop to initialize list) based on the number of nodes requested:

Write a function called listSize that takes two input parameters - the pointer head OR tail plus a parameter of which direction to traverse - to traverse the linked list and return the number of elements in the list. Write another function printForwardElem that prints out the value of every element of the linked list in sequence starting from head forward to the end of the list. Write a third function printBackwardElem that prints out the value of every element of the linked list in sequence starting from tail backward to the beginning of the list. Design your driver to test these three functions.

Answer 1

Here is the solution for above code:

int listSize (struct listrec* item,String direction)
{
  struct listrec* cur = item;
  int size = 0;
  if(direction == "right"){

  while (cur != null)
  {
    ++size;
    cur = cur->Next;
  }
}
else {

while (cur != null)
  {
    ++size;
    cur = cur->prev;
  }

}
  return size;
}

void printForwardElem(struct listrec* item) {

   struct node *ptr = item;

   printf("\n[head] <=>");
   //start from the beginning
   while(ptr != NULL) {        
      printf(" %d <=>",ptr->data);
      ptr = ptr->next;
   }

   printf(" [last]\n");
}

void printBackwardElem(struct listrec*start)
{
    struct listrec*current = start;  //current points to first node
    if(current==NULL)  //if empty list, return 
       return;

    //now we are sure that atleast one node exists
    while(current->link != NULL) //goes until current == last node
    {
        current = current->link; //keep on going forward till end of list
    }

    //start from last node and keep going back till you cross the first node
    while(current != NULL)
    {
        DisplayNode((struct inventory*)current->item.value);
        current = current->prev; //go one node back
    }
}