Invertir una cola usando otra cola

Dada una cola . La tarea es invertir la cola utilizando otra cola vacía.

Ejemplos: 

Input: queue[] = {1, 2, 3, 4, 5}
Output: 5 4 3 2 1

Input: queue[] = {10, 20, 30, 40}
Output: 40 30 20 10

Acercarse:  

• Dada una cola y una cola vacía.
• El último elemento de la cola debe ser el primer elemento de la nueva cola.
• Para obtener el último elemento, es necesario abrir la cola uno por uno y agregarlo al final de la cola, tamaño: 1 vez.
• Entonces, después de eso, obtendremos el último elemento al frente de la cola. Ahora saque ese elemento y agréguelo a la nueva cola. Repita los pasos s – 1 veces donde s es el tamaño original de la cola.

A continuación se muestra la implementación del enfoque: 

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the reversed queue
queue<int> reverse(queue<int> q)
{
    // Size of queue
    int s = q.size();
 
    // Second queue
    queue<int> ans;
 
    for (int i = 0; i < s; i++) {
 
        // Get the last element to the
        // front of queue
        for (int j = 0; j < q.size() - 1; j++) {
            int x = q.front();
            q.pop();
            q.push(x);
        }
 
        // Get the last element and
        // add it to the new queue
        ans.push(q.front());
        q.pop();
    }
    return ans;
}
 
// Driver Code
int main()
{
    queue<int> q;
 
    // Insert elements
    q.push(1);
    q.push(2);
    q.push(3);
    q.push(4);
    q.push(5);
 
    q = reverse(q);
 
    // Print the queue
    while (!q.empty()) {
        cout << q.front() << " ";
        q.pop();
    }
 
    return 0;
}

// Java implementation of the above approach
import java.util.*;
class GFG
{
 
// Function to return the reversed queue
static Queue<Integer> reverse(Queue<Integer> q)
{
    // Size of queue
    int s = q.size();
 
    // Second queue
    Queue<Integer> ans = new LinkedList<>();
 
    for (int i = 0; i < s; i++)
    {
 
        // Get the last element to the
        // front of queue
        for (int j = 0; j < q.size() - 1; j++)
        {
            int x = q.peek();
            q.remove();
            q.add(x);
        }
 
        // Get the last element and
        // add it to the new queue
        ans.add(q.peek());
        q.remove();
    }
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
    Queue<Integer> q = new LinkedList<>();
 
    // Insert elements
    q.add(1);
    q.add(2);
    q.add(3);
    q.add(4);
    q.add(5);
 
    q = reverse(q);
 
    // Print the queue
    while (!q.isEmpty())
    {
        System.out.print(q.peek() + " ");
        q.remove();
    }
    }
}
 
// This code is contributed by Princi Singh

# Python3 implementation of the above approach
from collections import deque
 
# Function to return the reversed queue
def reverse(q):
     
    # Size of queue
    s = len(q)
 
    # Second queue
    ans = deque()
 
    for i in range(s):
 
        # Get the last element to the
        # front of queue
        for j in range(s - 1):
            x = q.popleft()
            q.appendleft(x)
 
        # Get the last element and
        # add it to the new queue
        ans.appendleft(q.popleft())
    return ans
 
# Driver Code
q = deque()
 
# Insert elements
q.append(1)
q.append(2)
q.append(3)
q.append(4)
q.append(5)
 
q = reverse(q)
 
# Print the queue
while (len(q) > 0):
    print(q.popleft(), end = " ")
 
# This code is contributed by Mohit Kumar

// C# Program to print the given pattern
using System;
using System.Collections.Generic;
     
class GFG
{
 
// Function to return the reversed queue
static Queue<int> reverse(Queue<int> q)
{
    // Size of queue
    int s = q.Count;
 
    // Second queue
    Queue<int> ans = new Queue<int>();
 
    for (int i = 0; i < s; i++)
    {
 
        // Get the last element to the
        // front of queue
        for (int j = 0; j < q.Count - 1; j++)
        {
            int x = q.Peek();
            q.Dequeue();
            q.Enqueue(x);
        }
 
        // Get the last element and
        // add it to the new queue
        ans.Enqueue(q.Peek());
        q.Dequeue();
    }
    return ans;
}
 
// Driver Code
public static void Main(String[] args)
{
    Queue<int> q = new Queue<int>();
 
    // Insert elements
    q.Enqueue(1);
    q.Enqueue(2);
    q.Enqueue(3);
    q.Enqueue(4);
    q.Enqueue(5);
 
    q = reverse(q);
 
    // Print the queue
    while (q.Count!=0)
    {
        Console.Write(q.Peek() + " ");
        q.Dequeue();
    }
    }
}
 
// This code is contributed by Princi Singh

<script>
 
// Javascript implementation of the above approach
 
// Function to return the reversed queue
function reverse(q)
{
     
    // Size of queue
    let s = q.length;
   
    // Second queue
    let ans = [];
   
    for(let i = 0; i < s; i++)
    {
         
        // Get the last element to the
        // front of queue
        for(let j = 0; j < q.length - 1; j++)
        {
            let x = q.shift();
            q.push(x);
        }
   
        // Get the last element and
        // add it to the new queue
        ans.push(q[0]);
        q.shift();
    }
    return ans;
}
 
// Driver Code
let q = [];
 
// Insert elements
q.push(1);
q.push(2);
q.push(3);
q.push(4);
q.push(5);
 
q = reverse(q);
 
// Print the queue
while (q.length != 0)
{
    document.write(q[0] + " ");
    q.shift();
}
 
// This code is contributed by patel2127
 
</script>

5 4 3 2 1

Complejidad temporal: O(n ² ) 

Espacio Auxiliar: O(n)