Implement a Stack using Linked Lists

Here comes the fun part or i should say the extensive part. We can learn one data structure and then mess with it to create other data structures, here we are going to see how to implement a Stack using a singly linked list. 

For this we must keep in mind that a stack follows Last in First out structure (LIFO), Now we are ready to tackle this problem with certain Stack operations like Push, pop, peek-to give top element and display in mind.

Code-->

Class Node{
    int data;
    Node* link;
};

Node* top;

//function to add element 
void push(int data)
{
        Node* temp= new Node(data);
        if(!temp)
        {
                cout<<"Overflow";
                exit(1);
        }    

    temp->link=top;
    top=temp;
}

int empty()
{
    return top==NULL;
}

int peek()
{
    if(!empty())
        return top->data;
    else
        exit(1);
}
//function to remove an element

void pop()
{
        Node* temp;
        if(top==NULL)
                cout<<"Underflow";
                exit(1);
        else
            temp=top;
            top=top->link;
            temp->link=NULL;
            free(temp);
}

//function to display elements
void display()
{
        Node* temp;
        if(top==NULL)
            cout<<"underflow";
            exit(1);
        else
            temp=top;
            while(temp!=NULL)
                cout<<temp->data;    
                temp=temp->link;
}




Popular posts from this blog

Finding the Subarrays (Hackerearth)

Image
Finding subarrays can become a cumbersome task to handle if we don't know the right approach for it. In this particular question we will be using a new method.  Pair is used to combine together two values which may be different in type. Pair provides a way to store two heterogeneous objects as a single unit. Such as vector<first attribute, second attribute>. To access these attributes we use keywords 'first' and 'second' Approach - Input values and find total sum of all terms Create a loop and calculate average term by term, use logic ->  sum= first element this_average=sum/elements in the sum SZ= Elements left in array= size-elements in sum other_average=(total-sum)/SZ Use if loop to find whether this_average is bigger than other_average or not. If true, input the (i+1)th and (j+1)th element in the pair, Increment counter by 1 Sort the pair and output it. Code- #include< bits/stdc++.h > using namespace std ; int main () {       ...
Read more

Palindrome Index (Hackerrank)

Image
Here we have another Hackerrank question on String manipulation, this one involving the classic Palindrome index.  Question-  Approach - Pretty simple, we just need to check values one by one, iterating a loop from index 0 being compared to index n, where n is the last value in the index, narrowing it down to inner values one by one. The point where the elements do not stay equal is the point where element needs to be removed in order to make the string into a palindrome  The steps are -  initialise i = 0; j= string.length()-1 //last value in the string Run a loop from i to middle of the string, i increments while j decrements, bringing us a section to focus on within the string Run an if condition which checks whether character at i is equal to j or not If it is not equal, we check if the element after character at i is equal to j or not If yes, we check whether the element before j is equal to character at i and character after i is subsequently equal to element at...
Read more

Sherlock and Array (Hackerrank)

Image
Sometimes it's better to stop and first formulate a way to solve a problem before jumping right into it. Why? Well because it gives you more time to better understand a problem and secondly, in some cases even a simpler way to solve a problem.  Remember kids! When solving algorithms, Math is your friend! I realized that a little late into this question and when i did, it was just a 5 min snippet. Question-  Approach - It's quite simple and straight forward.  Instead of dividing array into two halves, getting their sums, checking whether the sums are equal and trying to find out if an element which causes sum to be unequal exist or not, here's a better approach. array - [5,6,8,11] , here we have 5+6=11 in LHS and 11 in RHS with 8 being an extra element. Imagine LHS and RHS as same, ex - x+y+x= sum. i.e 2x= sum-y. Here we just need to check if this equality holds, if it does we print YES and add value to x, else we print NO Code- 
Read more