Friday, July 20, 2007

Introduction to Linked Stacks

In the article Data Structures: Introduction to Stacks, we saw that there was one major disadvantage of representing stacks using arrays- the stack like the array could have a limited number of elements, while stacks should be able to grow up to any number of elements. Besides this there were other disadvantages too.

In one of the other article about Linked Lists, we noticed one useful property of linked lists that they can grow up to any size to accommodate for the addition of elements and it efficiently uses the memory too.

So if we combine both of this to from a linked version of the stack then it won’t have the shortcomings that the array version had.

This is what this article is all about.

pushing and popping

As you know that addition of elements to the stack is known as pushing while retrieval is known as popping.

The process of pushing and popping in case of linked version of stack is slightly different from the array version. The following graphics will clear it though!

pushing of elements in the linked stack

FIG: pushing of elements in the linked stack

popping of the elements from the linked stack

FIG: popping of the elements from the linked stack

Now that you know how the basic operations are performed on linked stacks I present you with the example program to illustrate this.

As always I would strongly recommend you to read the comments!

  // -- Linked stacks --
  // Example program to 
  // illustrate basic
  // push and pop to a
  // linked stack

  // node class, this will
  // represent the nodes or elements
  // of the linked stacks
  class node
     int info;
     node *link;

  // declare global objects
  node *start=NULL;

  // function prototypes
  void push(int);
  int pop();

  void main(void)
   int ch=0,num;

    cout<<"1> Push";
    cout<<"\n2> Pop";
    cout<<"\n3> Quit\n";


     case 1:
     cout<<"enter element:";


     case 2:
     cout<<"\n\nPopped: ";

  // below is a bit confusing
  // part.
  // here all the nodes that
  // we have allocated are
  // being freed up
  node temp;
   // store the next node
   // to the one being deleted
   // delete the node
   delete start;

   // retrieve the next node
   // to be deleted;

  // pushes an element 'inf'
  // to the linked stack
  void push(int inf)
   node *temp1;
   node temp2;

   // if the element to be added
   // is the first element of 
   // linked stack
    // allocate a new node
    temp1=new node;

    // make start point at it
   // if not
    // store the information
    // about the node pointed
    // by 'start'>info;;

    temp1=new node;
    // insert the new node
    // at the beginning
    // and make its link
    // point to the prev.
    // node pointed by 'start'


  // returns an element from the
  // linked stack
  int pop()
   node temp;

    // store info. about
    // the first element
    // that has to be pooped

    // delete the node
    delete start;
    // make start point at the
    // next node which had been
    // stored;


   return NULL;


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1 comment:

  1. Thanks for the information. very informative article.


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