Calculating powers of integers

Question

Is there any other way in Java to calculate a power of an integer?

I use Math.pow(a, b) now, but it returns a double, and that is usually a

Answer 1

import java.util.Scanner;

class Solution {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();

        for (int i = 0; i < t; i++) {

            try {
                long x = sc.nextLong();
                System.out.println(x + " can be fitted in:");
                if (x >= -128 && x <= 127) {
                    System.out.println("* byte");
                }
                if (x >= -32768 && x <= 32767) {
                    //Complete the code
                    System.out.println("* short");
                    System.out.println("* int");
                    System.out.println("* long");
                } else if (x >= -Math.pow(2, 31) && x <= Math.pow(2, 31) - 1) {
                    System.out.println("* int");
                    System.out.println("* long");
                } else {
                    System.out.println("* long");
                }
            } catch (Exception e) {
                System.out.println(sc.next() + " can't be fitted anywhere.");
            }

        }
    }
}

Answer 2

I managed to modify(boundaries, even check, negative nums check) Qx__ answer. Use at your own risk. 0^-1, 0^-2 etc.. returns 0.

private static int pow(int x, int n) {
        if (n == 0)
            return 1;
        if (n == 1)
            return x;
        if (n < 0) { // always 1^xx = 1 && 2^-1 (=0.5 --> ~ 1 )
            if (x == 1 || (x == 2 && n == -1))
                return 1;
            else
                return 0;
        }
        if ((n & 1) == 0) { //is even 
            long num = pow(x * x, n / 2);
            if (num > Integer.MAX_VALUE) //check bounds
                return Integer.MAX_VALUE; 
            return (int) num;
        } else {
            long num = x * pow(x * x, n / 2);
            if (num > Integer.MAX_VALUE) //check bounds
                return Integer.MAX_VALUE;
            return (int) num;
        }
    }

Answer 3

When it's power of 2. Take in mind, that you can use simple and fast shift expression 1 << exponent

example:

2² = 1 << 2 = (int) Math.pow(2, 2)
2¹⁰ = 1 << 10 = (int) Math.pow(2, 10)

For larger exponents (over 31) use long instead

2³² = 1L << 32 = (long) Math.pow(2, 32)

btw. in Kotlin you have shl instead of << so

(java) 1L << 32 = 1L shl 32 (kotlin)

Answer 4

Google Guava has math utilities for integers. IntMath

Answer 5

A simple (no checks for overflow or for validity of arguments) implementation for the repeated-squaring algorithm for computing the power:

/** Compute a**p, assume result fits in a 32-bit signed integer */ 
int pow(int a, int p)
{
    int res = 1;
    int i1 = 31 - Integer.numberOfLeadingZeros(p); // highest bit index
    for (int i = i1; i >= 0; --i) {
        res *= res;
        if ((p & (1<<i)) > 0)
            res *= a;
    }
    return res;
}

The time complexity is logarithmic to exponent p (i.e. linear to the number of bits required to represent p).

Answer 6

There some issues with pow method:

1. We can replace (y & 1) == 0; with y % 2 == 0
   bitwise operations always are faster.

Your code always decrements y and performs extra multiplication, including the cases when y is even. It's better to put this part into else clause.

public static long pow(long x, int y) {
        long result = 1;
        while (y > 0) {
            if ((y & 1) == 0) {
                x *= x;
                y >>>= 1;
            } else {
                result *= x;
                y--;
            }
        }
        return result;
    }