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Raise x to the power n using Recursion in Java
The power of a number can be calculated using recursion. Here the number is x and it is raised to power n.
A program that demonstrates this is given as follows:
Example
public class Demo {
static double pow(double x, int n) {
if (n != 0)
return (x * pow(x, n - 1));
else
return 1;
}
public static void main(String[] args) {
System.out.println("7 to the power 3 is " + pow(7, 3));
System.out.println("4 to the power 1 is " + pow(4, 1));
System.out.println("9 to the power 0 is " + pow(9, 0));
}
}Output
7 to the power 3 is 343.0 4 to the power 1 is 4.0 9 to the power 0 is 1.0
Now let us understand the above program.
The method pow() calculates x raised to the power n. If n is not 0, it recursively calls itself and returns x * pow(x, n - 1). If n is 0, it returns 1. A code snippet which demonstrates this is as follows:
static double pow(double x, int n) {
if (n != 0)
return (x * pow(x, n - 1));
else
return 1;
}In main(), the method pow() is called with different values. A code snippet which demonstrates this is as follows:
public static void main(String[] args) {
System.out.println("7 to the power 3 is " + pow(7, 3));
System.out.println("4 to the power 1 is " + pow(4, 1));
System.out.println("9 to the power 0 is " + pow(9, 0));
}
Updated on: 30-Jul-2019
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