Implementar una pila usando una sola cola

Recibimos una estructura de datos de cola, la tarea es implementar la pila utilizando solo la estructura de datos de cola dada.
Hemos discutido una solución que usa dos colas . En este artículo, se analiza una nueva solución que utiliza solo una cola. Esta solución asume que podemos encontrar el tamaño de la cola en cualquier punto. La idea es mantener el elemento recién insertado siempre al frente de la cola, manteniendo el mismo orden de los elementos anteriores. 

A continuación se muestran los pasos completos.

C++

// C++ program to implement a stack using
// single queue
#include<bits/stdc++.h>
using namespace std;
 
// User defined stack that uses a queue
class Stack
{
    queue<int>q;
public:
    void push(int val);
    void pop();
    int top();
    bool empty();
};
 
// Push operation
void Stack::push(int val)
{
    //  Get previous size of queue
    int s = q.size();
 
    // Push current element
    q.push(val);
 
    // Pop (or Dequeue) all previous
    // elements and put them after current
    // element
    for (int i=0; i<s; i++)
    {
        // this will add front element into
        // rear of queue
        q.push(q.front());
 
        // this will delete front element
        q.pop();
    }
}
 
// Removes the top element
void Stack::pop()
{
    if (q.empty())
        cout << "No elements\n";
    else
        q.pop();
}
 
// Returns top of stack
int  Stack::top()
{
    return (q.empty())? -1 : q.front();
}
 
// Returns true if Stack is empty else false
bool Stack::empty()
{
    return (q.empty());
}
 
// Driver code
int main()
{
    Stack s;
    s.push(10);
    s.push(20);
    cout << s.top() << endl;
    s.pop();
    s.push(30);
    s.pop();
    cout << s.top() << endl;
    return 0;
}

Java

// Java program to implement stack using a
// single queue
 
import java.util.LinkedList;
import java.util.Queue;
 
public class stack
{
    Queue<Integer> q = new LinkedList<Integer>();
     
    // Push operation
    void push(int val)
    {
        // get previous size of queue
        int size = q.size();
         
        // Add current element
        q.add(val);
         
        // Pop (or Dequeue) all previous
        // elements and put them after current
        // element
        for (int i = 0; i < size; i++)
        {
            // this will add front element into
            // rear of queue
            int x = q.remove();
            q.add(x);
        }
    }
     
    // Removes the top element
    int pop()
    {
        if (q.isEmpty())
        {
            System.out.println("No elements");
            return -1;
        }
        int x = q.remove();
        return x;
    }
     
    // Returns top of stack
    int top()
    {
        if (q.isEmpty())
            return -1;
        return q.peek();
    }
     
    // Returns true if Stack is empty else false
    boolean isEmpty()
    {
        return q.isEmpty();
    }
 
    // Driver program to test above methods
    public static void main(String[] args)
    {
        stack s = new stack();
        s.push(10);
        s.push(20);
        System.out.println("Top element :" + s.top());
        s.pop();
        s.push(30);
        s.pop();
        System.out.println("Top element :" + s.top());
    }
}
 
// This code is contributed by Rishabh Mahrsee

Python3

# Python3 program to implement stack using a
# single queue
  
q = []
 
# append operation
def append(val):
 
    # get previous size of queue
    size = len(q)
 
    # Add current element
    q.append(val);
 
    # Pop (or Dequeue) all previous
    # elements and put them after current
    # element
    for i in range(size):
 
        # this will add front element into
        # rear of queue
        x = q.pop(0);
        q.append(x);
            
# Removes the top element
def pop():
 
    if (len(q) == 0):
 
        print("No elements");
        return -1;
     
    x = q.pop(0);
    return x;
 
# Returns top of stack
def top():
 
    if(len(q) == 0):
        return -1;
    return q[-1]
 
# Returns true if Stack is empty else false
def isEmpty():
 
    return len(q)==0;
 
# Driver program to test above methods
if __name__=='__main__':
 
    s = []
 
    s.append(10);
    s.append(20);
    print("Top element :" + str(s[-1]));
    s.pop();
    s.append(30);
    s.pop();
    print("Top element :" + str(s[-1]));
     
    # This code is contributed by rutvik_56.

C#

// C# program to implement stack using a
// single queue
using System;
using System.Collections.Generic;
 
public class stack
{
    Queue<int> q = new Queue<int>();
     
    // Push operation
    void push(int val)
    {
        // get previous size of queue
        int size = q.Count;
         
        // Add current element
        q.Enqueue(val);
         
        // Pop (or Dequeue) all previous
        // elements and put them after current
        // element
        for (int i = 0; i < size; i++)
        {
            // this will add front element into
            // rear of queue
            int x = q.Dequeue();
            q.Enqueue(x);
        }
    }
     
    // Removes the top element
    int pop()
    {
        if (q.Count == 0)
        {
            Console.WriteLine("No elements");
            return -1;
        }
        int x = q.Dequeue();
        return x;
    }
     
    // Returns top of stack
    int top()
    {
        if (q.Count == 0)
            return -1;
        return q.Peek();
    }
     
    // Returns true if Stack is empty else false
    bool isEmpty()
    {
        if(q.Count == 0)
            return true;
        return false;
    }
 
    // Driver program to test above methods
    public static void Main(String[] args)
    {
        stack s = new stack();
        s.push(10);
        s.push(20);
        Console.WriteLine("Top element :" + s.top());
        s.pop();
        s.push(30);
        s.pop();
        Console.WriteLine("Top element :" + s.top());
    }
}
 
// This code has been contributed by Rajput-Ji

Javascript

<script>
    // Javascript program to implement stack using a single queue
    let q = [];
      
    // Push operation
    function Push(val)
    {
        // get previous size of queue
        let Size = q.length;
          
        // Add current element
        q.push(val);
          
        // Pop (or Dequeue) all previous
        // elements and put them after current
        // element
        for (let i = 0; i < Size; i++)
        {
            // this will add front element into
            // rear of queue
            let x = q[0];
            q.shift();
            q.push(x);
        }
    }
  
    // Removes the top element
    function Pop()
    {
        if (isEmpty())
        {
            document.write("No elements" + "</br>");
            return -1;
        }
        let x = q[0];
        q.shift();
        return x;
    }
      
    // Returns top of stack
    function Top()
    {
        if (isEmpty())
            return -1;
        return q[0];
    }
     
    // Returns true if Stack is empty else false
    function isEmpty()
    {
        if(q.length == 0)
            return true;
        return false;
    }
     
    Push(10);
    Push(20);
    document.write(Top() + "</br>");
    Pop();
    Push(30);
    Pop();
    document.write(Top() + "</br>");
 
// This code is contributed by decode2207.
</script>

Publicación traducida automáticamente

Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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