Imprima números pares e impares en un rango dado usando recursividad

Dados dos números enteros L y R , la tarea es imprimir todos los números pares e impares de L a R usando recursividad .

Ejemplos:

Entrada: L = 1, R = 10
Salida: 
Números pares: 2 4 6 8 10
Números impares: 1 3 5 7 9

Entrada: L = 10, R = 25 
Salida: 
Números pares: 10 12 14 16 18 20 22 24 
Números impares: 11 13 15 17 19 21 23 25

Enfoque: siga los pasos a continuación para resolver el problema usando Recursion :

• Atraviesa el rango [R, L] .
• Imprima los elementos impares del rango usando recursividad usando la siguiente relación de recurrencia:

Impar(L, R) = R % 2 == 1? Impar(L, R – 2) : Impar(L, R – 1) 

• Imprima los elementos pares del rango usando recursividad usando la siguiente relación de recurrencia :

Par(L, R) = R % 2 == 0 ? Par (L, R – 2) : Par (L, R – 1) 

A continuación se muestra la implementación del enfoque anterior:

// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print all the
// even numbers from L to R
void Even(int L, int R)
{
 
    // Base case
    if (R < L) {
        return;
    }
 
    // Recurrence relation
    R % 2 == 0 ? Even(L, R - 2)
               : Even(L, R - 1);
 
    // Check if R is even
    if (R % 2 == 0) {
        cout << R << " ";
    }
}
 
// Function to print all the
// odd numbers from L to R
void Odd(int L, int R)
{
 
    // Base case
    if (R < L) {
        return;
    }
 
    // Recurrence relation
    R % 2 == 1 ? Odd(L, R - 2)
               : Odd(L, R - 1);
 
    // Check if R is even
    if (R % 2 == 1) {
        cout << R << " ";
    }
}
 
// Driver Code
int main()
{
    int L = 10, R = 25;
    cout << "Even numbers:";
 
    // Print all the
    // even numbers
    Even(L, R);
    cout << endl;
 
    // Print all the
    // odd numbers
    cout << "Odd numbers:";
    Odd(L, R);
}

// Java program to implement
// the above approach
import java.util.*;
class GFG{
 
// Function to print
// all the even numbers
// from L to R
static void Even(int L,
                 int R)
{
  // Base case
  if (R < L)
  {
    return;
  }
 
  // Recurrence relation
  if(R % 2 == 0 )
    Even(L, R - 2);
  else
    Even(L, R - 1);
 
  // Check if R is even
  if (R % 2 == 0)
  {
    System.out.print(R + " ");
  }
}
 
// Function to print
// all the odd numbers
// from L to R
static void Odd(int L,
                int R)
{
  // Base case
  if (R < L)
  {
    return;
  }
 
  // Recurrence relation
  if(R % 2 == 1 )
    Odd(L, R - 2);
  else
    Odd(L, R - 1);
 
  // Check if R is even
  if (R % 2 == 1)
  {
    System.out.print(R + " ");
  }
}
 
// Driver Code
public static void main(String[] args)
{
  int L = 10, R = 25;
  System.out.print("Even numbers:");
 
  // Print all the
  // even numbers
  Even(L, R);
  System.out.println();
 
  // Print all the
  // odd numbers
  System.out.print("Odd numbers:");
  Odd(L, R);
}
}
 
// This code is contributed by Rajput-Ji

# Python3 program to implement
# the above approach
 
# Function to print all the
# even numbers from L to R
def Even(L, R):
     
    # Base case
    if (R < L):
        return
 
    # Recurrence relation
    if (R % 2 == 0):
        Even(L, R - 2)
    else:
        Even(L, R - 1)
 
    # Check if R is even
    if (R % 2 == 0):
        print(R, end = " ")
 
# Function to print all the
# odd numbers from L to R
def Odd(L, R):
     
    # Base case
    if (R < L):
        return
 
    # Recurrence relation
    if (R % 2 == 1):
        Odd(L, R - 2)
    else:
        Odd(L, R - 1)
 
    # Check if R is even
    if (R % 2 == 1):
        print(R, end = " ")
 
# Driver Code
if __name__ == '__main__':
     
    L = 10
    R = 25
     
    print("Even numbers:")
 
    # Print all the
    # even numbers
    Even(L, R)
    print()
 
    # Print all the
    # odd numbers
    print("Odd numbers:")
    Odd(L, R)
 
# This code is contributed by Amit Katiyar

// C# program to implement
// the above approach
using System;
class GFG{
 
// Function to print
// all the even numbers
// from L to R
static void Even(int L,
                 int R)
{
  // Base case
  if (R < L)
  {
    return;
  }
 
  // Recurrence relation
  if(R % 2 == 0 )
    Even(L, R - 2);
  else
    Even(L, R - 1);
 
  // Check if R is even
  if (R % 2 == 0)
  {
    Console.Write(R + " ");
  }
}
 
// Function to print
// all the odd numbers
// from L to R
static void Odd(int L,
                int R)
{
  // Base case
  if (R < L)
  {
    return;
  }
 
  // Recurrence relation
  if(R % 2 == 1 )
    Odd(L, R - 2);
  else
    Odd(L, R - 1);
 
  // Check if R is even
  if (R % 2 == 1)
  {
    Console.Write(R + " ");
  }
}
 
// Driver Code
public static void Main(String[] args)
{
  int L = 10, R = 25;
  Console.Write("Even numbers:");
 
  // Print all the
  // even numbers
  Even(L, R);
  Console.WriteLine();
 
  // Print all the
  // odd numbers
  Console.Write("Odd numbers:");
  Odd(L, R);
}
}
 
// This code is contributed by 29AjayKumar

<script>
 
// Java script program to implement
// the above approach
 
// Function to print
// all the even numbers
// from L to R
function Even(L, R)
{
     
    // Base case
    if (R < L)
    {
        return;
    }
     
    // Recurrence relation
    if (R % 2 == 0 )
    {
        Even(L, R - 2);
    }
    else
    {
        Even(L, R - 1);
    }
     
    // Check if R is even
    if (R % 2 == 0)
    {
        document.write(R + " ");
    }
}
 
// Function to print
// all the odd numbers
// from L to R
function Odd(L, R)
{
     
    // Base case
    if (R < L)
    {
        return;
    }
     
    // Recurrence relation
    if (R % 2 == 1 )
    {
        Odd(L, R - 2);
    }
    else
    {
        Odd(L, R - 1);
    }
     
    // Check if R is even
    if (R % 2 == 1)
    {
        document.write(R + " ");
    }
}
 
// Driver Code
let L = 10;
let R = 25;
document.write("Even numbers:");
 
// Print all the
// even numbers
Even(L, R);
document.write("<br>");
 
// Print all the
// odd numbers
document.write("Odd numbers:");
Odd(L, R);
 
// This code is contributed by sravan kumar G
 
</script>

Even numbers:10 12 14 16 18 20 22 24 
Odd numbers:11 13 15 17 19 21 23 25

Complejidad temporal: O(N)
Espacio auxiliar: O(1)