Power of a Number using Recursion

Easy Recursion

Compute base raised to the power exp using recursion.

Sample input

base = 2, exp = 10

Sample output

1024

Solution

def power(base, exp):
    if exp == 0:
        return 1
    return base * power(base, exp - 1)

print(power(2, 10))

function power(base, exp) {
  if (exp === 0) return 1;
  return base * power(base, exp - 1);
}

console.log(power(2, 10));

public class Main {
    static long power(int base, int exp) {
        if (exp == 0) return 1;
        return base * power(base, exp - 1);
    }

    public static void main(String[] args) {
        System.out.println(power(2, 10));
    }
}

fun power(base: Int, exp: Int): Long {
    if (exp == 0) return 1
    return base * power(base, exp - 1)
}

fun main() {
    println(power(2, 10))
}

func power(_ base: Int, _ exp: Int) -> Int {
    if exp == 0 {
        return 1
    }
    return base * power(base, exp - 1)
}

print(power(2, 10))

int power(int base, int exp) {
  if (exp == 0) return 1;
  return base * power(base, exp - 1);
}

void main() {
  print(power(2, 10));
}

#include <iostream>
using namespace std;

long long power(int base, int exp) {
    if (exp == 0) return 1;
    return base * power(base, exp - 1);
}

int main() {
    cout << power(2, 10) << endl;
    return 0;
}

#include <stdio.h>

long long power(int base, int exp) {
    if (exp == 0) return 1;
    return base * power(base, exp - 1);
}

int main() {
    printf("%lld\n", power(2, 10));
    return 0;
}