Dados dos números N y M. La tarea es encontrar el producto de los 2 números usando recursividad.
Nota : Los números pueden ser tanto positivos como negativos.
Ejemplos :
Input : N = 5 , M = 3 Output : 15 Input : N = 5 , M = -3 Output : -15 Input : N = -5 , M = 3 Output : -15 Input : N = -5 , M = -3 Output:15
Una solución recursiva al problema anterior solo para números positivos ya se discutió en el artículo anterior . En esta publicación, se analiza una solución recursiva para encontrar el producto de números positivos y negativos.
A continuación se muestra el enfoque paso a paso:
- Compruebe si uno o ambos números son negativos.
- Si el número pasado en el segundo parámetro es negativo, intercambie los parámetros y vuelva a llamar a la función.
- Si ambos parámetros son negativos, vuelva a llamar a la función y pase los valores absolutos de los números como parámetros.
- Si n>m llame a la función con parámetros intercambiados para reducir el tiempo de ejecución de la función.
- Siempre que m no sea 0, siga llamando a la función con el subcaso n, m-1 y devuelva n + multrecur(n, m-1).
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to find product of two numbers
// using recursion
#include <iostream>
using namespace std;
// Recursive function to calculate the product
// of 2 integers
int multrecur(int n, int m)
{
// case 1 : n<0 and m>0
// swap the position of n and m to keep second
// parameter positive
if (n > 0 && m < 0) {
return multrecur(m, n);
}
// case 2 : both n and m are less than 0
// return the product of their absolute values
else if (n < 0 && m < 0) {
return multrecur((-1 * n), (-1 * m));
}
// if n>m , swap n and m so that recursion
// takes less time
if (n > m) {
return multrecur(m, n);
}
// as long as m is not 0 recursively call multrecur for
// n and m-1 return sum of n and the product of n times m-1
else if (m != 0) {
return n + multrecur(n, m - 1);
}
// m=0 then return 0
else {
return 0;
}
}
// Driver code
int main()
{
cout << "5 * 3 = " << multrecur(5, 3) << endl;
cout << "5 * (-3) = " << multrecur(5, -3) << endl;
cout << "(-5) * 3 = " << multrecur(-5, 3) << endl;
cout << "(-5) * (-3) = " << multrecur(-5, -3) << endl;
return 0;
}
Java
//Java program to find product of two numbers
//using recursion
public class GFG {
//Recursive function to calculate the product
//of 2 integers
static int multrecur(int n, int m)
{
// case 1 : n<0 and m>0
// swap the position of n and m to keep second
// parameter positive
if (n > 0 && m < 0) {
return multrecur(m, n);
}
// case 2 : both n and m are less than 0
// return the product of their absolute values
else if (n < 0 && m < 0) {
return multrecur((-1 * n), (-1 * m));
}
// if n>m , swap n and m so that recursion
// takes less time
if (n > m) {
return multrecur(m, n);
}
// as long as m is not 0 recursively call multrecur for
// n and m-1 return sum of n and the product of n times m-1
else if (m != 0) {
return n + multrecur(n, m - 1);
}
// m=0 then return 0
else {
return 0;
}
}
//Driver code
public static void main(String[] args) {
System.out.println("5 * 3 = " + multrecur(5, 3));
System.out.println("5 * (-3) = " + multrecur(5, -3));
System.out.println("(-5) * 3 = " + multrecur(-5, 3));
System.out.println("(-5) * (-3) = " +multrecur(-5, -3));
}
}
Python3
# Python 3 program to find product of two numbers
# using recursion
# Recursive function to calculate the product
# of 2 integers
def multrecur(n, m) :
# case 1 : n<0 and m>0
# swap the position of n and m to keep second
# parameter positive
if n > 0 and m < 0 :
return multrecur(m,n)
# case 2 : both n and m are less than 0
# return the product of their absolute values
elif n < 0 and m < 0 :
return multrecur((-1 * n),(-1 * m))
# if n>m , swap n and m so that recursion
# takes less time
if n > m :
return multrecur(m, n)
# as long as m is not 0 recursively call multrecur for
# n and m-1 return sum of n and the product of n times m-1
elif m != 0 :
return n + multrecur(n, m-1)
# m=0 then return 0
else :
return 0
# Driver Code
if __name__ == "__main__" :
print("5 * 3 =",multrecur(5, 3))
print("5 * (-3) =",multrecur(5, -3))
print("(-5) * 3 =",multrecur(-5, 3))
print("(-5) * (-3) =",multrecur(-5, -3))
# This code is contributed by ANKITRAI1
C#
// C# program to find product of
// two numbers using recursion
using System;
class GFG
{
// Recursive function to calculate
// the product of 2 integers
static int multrecur(int n, int m)
{
// case 1 : n<0 and m>0
// swap the position of n and m
// to keep second parameter positive
if (n > 0 && m < 0)
{
return multrecur(m, n);
}
// case 2 : both n and m are less than 0
// return the product of their absolute values
else if (n < 0 && m < 0)
{
return multrecur((-1 * n), (-1 * m));
}
// if n>m , swap n and m so that
// recursion takes less time
if (n > m)
{
return multrecur(m, n);
}
// as long as m is not 0 recursively
// call multrecur for n and m-1 return
// sum of n and the product of n times m-1
else if (m != 0)
{
return n + multrecur(n, m - 1);
}
// m=0 then return 0
else
{
return 0;
}
}
// Driver code
public static void Main()
{
Console.WriteLine("5 * 3 = " +
multrecur(5, 3));
Console.WriteLine("5 * (-3) = " +
multrecur(5, -3));
Console.WriteLine("(-5) * 3 = " +
multrecur(-5, 3));
Console.WriteLine("(-5) * (-3) = " +
multrecur(-5, -3));
}
}
// This code is contributed by anuj_67
PHP
<?php
// PHP program to find product of
// two numbers using recursion
// Recursive function to calculate
// the product of 2 integers
function multrecur($n, $m)
{
// case 1 : n<0 and m>0
// swap the position of n and m to keep second
// parameter positive
if ($n > 0 && $m < 0)
{
return multrecur($m, $n);
}
// case 2 : both n and m are less than 0
// return the product of their absolute values
else if ($n < 0 && $m < 0)
{
return multrecur((-1 * $n),
(-1 * $m));
}
// if n>m , swap n and m so that
// recursion takes less time
if ($n > $m)
{
return multrecur($m, $n);
}
// as long as m is not 0 recursively call multrecur for
// n and m-1 return sum of n and the product of n times m-1
else if ($m != 0)
{
return $n + multrecur($n, $m - 1);
}
// m=0 then return 0
else
{
return 0;
}
}
// Driver code
echo "5 * 3 = " . multrecur(5, 3) . "\n";
echo "5 * (-3) = " . multrecur(5, -3) . "\n";
echo "(-5) * 3 = " . multrecur(-5, 3) . "\n";
echo "(-5) * (-3) = " . multrecur(-5, -3) . "\n";
// This code is contributed by mits
?>
Javascript
<script>
// Javascript program to find product
// of two numbers using recursion
// Recursive function to calculate the
// product of 2 integers
function multrecur(n, m)
{
// case 1 : n<0 and m>0
// Swap the position of n and m to
// keep second parameter positive
if (n > 0 && m < 0)
{
return multrecur(m, n);
}
// case 2 : both n and m are less than 0
// return the product of their absolute values
else if (n < 0 && m < 0)
{
return multrecur((-1 * n), (-1 * m));
}
// If n>m , swap n and m so that recursion
// takes less time
if (n > m)
{
return multrecur(m, n);
}
// As long as m is not 0 recursively
// call multrecur for n and m-1
// return sum of n and the product
// of n times m-1
else if (m != 0)
{
return n + multrecur(n, m - 1);
}
// m=0 then return 0
else
{
return 0;
}
}
// Driver code
document.write("5 * 3 = " + multrecur(5, 3) + "<br>");
document.write("5 * (-3) = " + multrecur(5, -3) + "<br>");
document.write("(-5) * 3 = " + multrecur(-5, 3) + "<br>");
document.write("(-5) * (-3) = " + multrecur(-5, -3) + "<br>");
// This code is contributed by rutvik_56
</script>
Producción:
5 * 3 = 15 5 * (-3) = -15 (-5) * 3 = -15 (-5) * (-3) = 15
Complejidad de tiempo: O(max(N, M))
Espacio auxiliar: O(max(N, M))
Publicación traducida automáticamente
Artículo escrito por neverSettle y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA