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stacks based on a linked list

Ryan Kinoti answered on July 13, 2022 Popularity 8/10 Helpfulness 1/10

answer stacks based on a linked list

stacks based on a linked list

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// Java program to Implement a stack
// using singly linked list
// import package
import static java.lang.System.exit;
  
// Create Stack Using Linked list
class StackUsingLinkedlist {
  
    // A linked list node
    private class Node {
  
        int data; // integer data
        Node link; // reference variable Node type
    }
    // create global top reference variable global
    Node top;
    // Constructor
    StackUsingLinkedlist()
    {
        this.top = null;
    }
  
    // Utility function to add an element x in the stack
    public void push(int x) // insert at the beginning
    {
        // create new node temp and allocate memory
        Node temp = new Node();
  
        // check if stack (heap) is full. Then inserting an
        //  element would lead to stack overflow
        if (temp == null) {
            System.out.print("\nHeap Overflow");
            return;
        }
  
        // initialize data into temp data field
        temp.data = x;
  
        // put top reference into temp link
        temp.link = top;
  
        // update top reference
        top = temp;
    }
  
    // Utility function to check if the stack is empty or not
    public boolean isEmpty()
    {
        return top == null;
    }
  
    // Utility function to return top element in a stack
    public int peek()
    {
        // check for empty stack
        if (!isEmpty()) {
            return top.data;
        }
        else {
            System.out.println("Stack is empty");
            return -1;
        }
    }
  
    // Utility function to pop top element from the stack
    public void pop() // remove at the beginning
    {
        // check for stack underflow
        if (top == null) {
            System.out.print("\nStack Underflow");
            return;
        }
  
        // update the top pointer to point to the next node
        top = (top).link;
    }
  
    public void display()
    {
        // check for stack underflow
        if (top == null) {
            System.out.printf("\nStack Underflow");
            exit(1);
        }
        else {
            Node temp = top;
            while (temp != null) {
  
                // print node data
                System.out.printf("%d->", temp.data);
  
                // assign temp link to temp
                temp = temp.link;
            }
        }
    }
}
// main class
public class GFG {
    public static void main(String[] args)
    {
        // create Object of Implementing class
        StackUsingLinkedlist obj = new StackUsingLinkedlist();
        // insert Stack value
        obj.push(11);
        obj.push(22);
        obj.push(33);
        obj.push(44);
  
        // print Stack elements
        obj.display();
  
        // print Top element of Stack
        System.out.printf("\nTop element is %d\n", obj.peek());
  
        // Delete top element of Stack
        obj.pop();
        obj.pop();
  
        // print Stack elements
        obj.display();
  
        // print Top element of Stack
        System.out.printf("\nTop element is %d\n", obj.peek());
    }
}

Popularity 8/10 Helpfulness 1/10 Language java

Source: www.geeksforgeeks.org

Tags: java linked-list

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Contributed on Jul 13 2022

Ryan Kinoti

0 Answers Avg Quality 2/10

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stacks based on a linked list

Comment

// Java program to Implement a stack
// using singly linked list
// import package
import static java.lang.System.exit;
  
// Create Stack Using Linked list
class StackUsingLinkedlist {
  
    // A linked list node
    private class Node {
  
        int data; // integer data
        Node link; // reference variable Node type
    }
    // create global top reference variable global
    Node top;
    // Constructor
    StackUsingLinkedlist()
    {
        this.top = null;
    }
  
    // Utility function to add an element x in the stack
    public void push(int x) // insert at the beginning
    {
        // create new node temp and allocate memory
        Node temp = new Node();
  
        // check if stack (heap) is full. Then inserting an
        //  element would lead to stack overflow
        if (temp == null) {
            System.out.print("\nHeap Overflow");
            return;
        }
  
        // initialize data into temp data field
        temp.data = x;
  
        // put top reference into temp link
        temp.link = top;
  
        // update top reference
        top = temp;
    }
  
    // Utility function to check if the stack is empty or not
    public boolean isEmpty()
    {
        return top == null;
    }
  
    // Utility function to return top element in a stack
    public int peek()
    {
        // check for empty stack
        if (!isEmpty()) {
            return top.data;
        }
        else {
            System.out.println("Stack is empty");
            return -1;
        }
    }
  
    // Utility function to pop top element from the stack
    public void pop() // remove at the beginning
    {
        // check for stack underflow
        if (top == null) {
            System.out.print("\nStack Underflow");
            return;
        }
  
        // update the top pointer to point to the next node
        top = (top).link;
    }
  
    public void display()
    {
        // check for stack underflow
        if (top == null) {
            System.out.printf("\nStack Underflow");
            exit(1);
        }
        else {
            Node temp = top;
            while (temp != null) {
  
                // print node data
                System.out.printf("%d->", temp.data);
  
                // assign temp link to temp
                temp = temp.link;
            }
        }
    }
}
// main class
public class GFG {
    public static void main(String[] args)
    {
        // create Object of Implementing class
        StackUsingLinkedlist obj = new StackUsingLinkedlist();
        // insert Stack value
        obj.push(11);
        obj.push(22);
        obj.push(33);
        obj.push(44);
  
        // print Stack elements
        obj.display();
  
        // print Top element of Stack
        System.out.printf("\nTop element is %d\n", obj.peek());
  
        // Delete top element of Stack
        obj.pop();
        obj.pop();
  
        // print Stack elements
        obj.display();
  
        // print Top element of Stack
        System.out.printf("\nTop element is %d\n", obj.peek());
    }
}

Popularity 8/10 Helpfulness 1/10 Language java

Source: www.geeksforgeeks.org

Tags: java linked-list linked-li

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Contributed on Jul 13 2022

Ryan Kinoti

0 Answers Avg Quality 2/10

Stack using linked list

Comment

#include<iostream>
using namespace std;
class node{
	public:
		int data;
		node* next;
		void push();
		void pop();
		void display();
};
node *top=NULL;
void node::push()
{
	node* newnode;
	newnode=new node();
	cout<<"Enter data : ";
	cin>>newnode->data;
	newnode->next=top;
	top=newnode;
}
void node::pop()
{
	if(top==NULL)
	{
		cout<<"Stack underflow!";
	}
	else
	{
		cout<<"poped element is : "<<top->data;
		top=top->next;
	}
}

void node::display()
{
	node *temp=top;
	if(temp==NULL)
	{
		cout<<"\nThe Stack is Empty!\n";
	}
	else
	{
	while(temp!=NULL)
	{
		cout<<temp->data<<" ";
		temp=temp->next;
	}
	}
}
int main()
{
	node obj;
 int ch;
   
    while(ch!=4){
    	cout<<"\n1) Push in stack"<<endl;
   		cout<<"2) Pop from stack"<<endl;
   		cout<<"3) Display stack"<<endl;
  		cout<<"4) Exit"<<endl;
    	cout<<"Enter choice: "<<endl;
      cin>>ch;
      switch(ch) {
         case 1: {
            obj.push();
            break;
         }
         case 2: {
            obj.pop();
            break;
         }
         case 3: {
            obj.display();
            break;
         }
         case 4: {
            exit(0);
            break;
         }
         default: {
            cout<<"Invalid Choice"<<endl;
         }
	
}
}
}

Popularity 7/10 Helpfulness 1/10 Language cpp

Source: Grepper

Tags: stack c++ linked-list using

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Contributed on Sep 06 2023

talha shahid

0 Answers Avg Quality 2/10

stacks based on a linked list

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