Learn javaScript
I recently started watching Essentials of javaScript by DoriSmith
Video tutorials are very easy to follow, productive and easy to grasp. The only problem I find with video tutorials is, when you get stuck or have some doubt few days after watching the tutorial, it’s hard to refer back. So, its always good to prepare some notes of important points which you can refer back easily. I’m going to share my notes here as I go along, I hope you find it use full.
Stacks are linear data structures. This means that their contexts are stored in what looks like a line (although vertically). This linear property, however, is not sufficient to discriminate a stack from other linear data structures. For example, an array is a sort of linear data structure in which you can access any element directly. In contrast, in a stack, you can only access the element at its top. In a stack Last thing In is the First thing Out. Thus, we say that a stack enforces LIFO order.
One disadvantage of implementing stack using an array is wastage of space. Most of the times most of the array is left unused. A better way to implement stack is by using a linked list. By using a single linked list to implement a stack there is no wastage of space.
Implementation of stack using a linked list
/* A simple implementation of linked list in C plus plus (CPP/C++) */ /* Compiler used: g++ */ #include<iostream> using namespace std; struct node{ int data; node *next; }; //This function pushes data(node) to the stack node * push(node*,node*); //This function pops data (node) and returns pointer to the poped node node * pop(node*); //This function shows the stack void show_stack(node*); ///These two global variables will help us track the start and end of the list node *f,*l; main() { int i,ch; f=l=NULL; cout<<"\t 1 to push"<<endl; cout<<"\t 2 to pop"<<endl; cout<<"\t 3 to show stack"<<endl; cout<<"\t 0 to exit"<<endl; while(1) { cout<<"Enter your choice:"; cin>>ch; switch(ch) { case 1: { if(f==NULL) l=f=push(f,l); else l=push(f,l); break; } case 2: { cout<<"popped:"<<pop(f)->data<<endl; break; } case 3: { show_stack(f); break; } case 0: break; default: cout<<"Please enter a proper choice"<<endl;break; } if(ch==0) break; } } node * push(node *f,node *l) { node *n; n=new node; cout<<"Enter data:"; cin>>n->data; n->next=NULL; if(f==NULL) {f=l=n;return f;} else { l->next=n; l=n;return l; } } void show_stack(node *f) { cout<<"showing data"<<endl; node *guest=f; while(guest!=NULL) { cout<<"\t"<<guest->data<<endl; guest=guest->next; } } node * pop(node *f) { node *guest,*lb; guest=lb=f; while(guest->next!=NULL) { lb=guest; guest=guest->next; } lb->next=NULL; l=lb; return guest; }Download the program: Implementation of stack using a single linked list
The simplest kind of linked list is a singly-linked list , which has one link per node. This link points to the next node in the list, or to a null value or empty list if it is the final node.A singly linked list’s node is divided into two parts. The first part holds or points to information about the node, and second part holds the address of next node. A singly linked list travels one way.
A singly-linked list containing two values: the value of the current node and a link to the next node
/* A simple implementation of linked list in C plus plus (CPP/C++) */ /* Program downloaded from www.GeeksPlanet.net For any help, post comment here http://geeksplanet.net/2008/11/data-structures/implementation-of-singly-linked-list-using-c-plus-plus/ Compiler used: g++ */ #include<iostream> using namespace std; struct node{ int data; node *next; }; /* This function adds a node at the end of the list and returns pointer to the added node, This takes pointer to the first node and pointer to the last node as argument. By giving the last node as argument we are saving some computer labour as it need not travel from the start to the end to find the last node */ node * addnode(node*,node*); /* This function just takes the first node as the argument and traverses the entire list displaying the data in each node. */ void shownodes(node*); /* This function takes the first node as argument and traverses the list until it finds the last node and deletes it and returns pointer to the new last node. */ node * delete_node(node*); /* *f and *l are the global variables keeping track of first and last nodes. */ node *f,*l; main() { int i,ch; f=l=NULL; cout<<"\t 1 to add a node"<<endl; cout<<"\t 2 to see the nodes"<<endl; cout<<"\t 3 to delete a node"<<endl; cout<<"\t 0 to exit"<<endl; while(1) { cout<<"Enter your choice:"; cin>>ch; switch(ch) { case 1: { if(f==NULL) l=f=addnode(f,l); //Since first node is the last node else l=addnode(f,l); //Last node is the new node added break; } case 2: shownodes(f);break; case 3: l=delete_node(f);break; // Last node has been changed to the last but one. case 0: break; default: cout<<"Please enter a proper choice"<<endl;break; } if(ch==0) break; } } node * addnode(node *f,node *l) { node *n; n=new node; //Allocating memory for the new node cout<<"Enter data:"; cin>>n->data; //This is going to be the last node, so its next point to NULL n->next=NULL; //If there is no first node, then the new node is the first and last node. if(f==NULL) {f=l=n;return f;} else { // Pointing the last node to the new node l->next=n; //Setting the new node as the last node l=n;return l; } } void shownodes(node *f) { cout<<"showing data"<<endl; node *guest=f; while(guest!=NULL) { cout<<"\t"<<guest->data<<endl; guest=guest->next; } } node * delete_node(node *f) { node *guest,*lb; guest=lb=f; while(guest->next!=NULL) { lb=guest; guest=guest->next; } lb->next=NULL; cout<<"node deleted"<<endl; return lb; }Download the prgoram linked-list.cpp
