原文:
给定两个数字 n 和 m，任务是用递归求这两个数字的乘积。 注:数字可以是正数，也可以是负数。
示例:
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
只有正数的上述问题的递归pg电子试玩链接的解决方案已经在中讨论过。在这篇文章中，讨论了求正数和负数乘积的递归解法。
下面是一步一步的方法:
- 检查一个或两个数字是否为负数。
- 如果在第二个参数中传递的数字是负数，交换参数并再次调用函数。
- 如果两个参数都是负数，则再次调用该函数，并将数字的绝对值作为参数传递。
- 如果 n>m，调用带有交换参数的函数，以减少函数的执行时间。
- 只要 m 不为 0，就继续用子基 n，m-1 调用函数，并返回 n multi recury(n，m-1)。
下面是上述方法的实现:
c
// c program to find product of two numbers
// using recursion
#include
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 语言(一种计算机语言，尤用于创建网站)
//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));
}
}
python 3
# 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
服务器端编程语言（professional hypertext preprocessor 的缩写）
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
?>
java 描述语言
output:
5 * 3 = 15
5 * (-3) = -15
(-5) * 3 = -15
(-5) * (-3) = 15
时间复杂度: o(max(n，m)) 辅助空间: o(max(n，m))
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