【Codewars】Folding your way to the moon

Codewars裏的 7kyu Kata。

題目說明：

Have you heard about the myth that if you fold a paper enough times, you can reach the moon with it? Sure you do, but exactly how many? Maybe it's time to write a program to figure it out.

You know that a piece of paper has a thickness of 0.0001m. Given distance in units of meters, calculate how many times you have to fold the paper to make the paper reach this distance.
(If you're not familiar with the concept of folding a paper: Each fold doubles its total thickness.)

Note: Of course you can't do half a fold. You should know what this means ;P

Also, if somebody is giving you a negative distance, it's clearly bogus and you should yell at them by returning null (or whatever equivalent in your language. In Shell please return None).

這一張厚度爲 0.0001m 的紙到目標值。求摺疊的次數。

解題代碼：

public class PaperFolder {
    public static Long fold(Double distance) {
        if (distance < 0)
            return null;
        if (distance < 0.0001)
            return new Long(0);

        double dis = Double.valueOf(distance);
        long res = 0;
        while (dis > 0.0001) {
            dis /= 2;
            res++;
        }
        return new Long(res);
    }
}

Test Cases:

import org.junit.Test;
import static org.junit.Assert.assertEquals;
import org.junit.runners.JUnit4;
import java.lang.Math;
import java.util.Random;

public class SolutionTest {

    public static Long solution(Double distance) {
        if (distance < 0)
            return null;
        if (distance < 0.0001)
            return Long.valueOf(0L);
        String str = Double.toString(Math.ceil(Math.log(distance * 10000) / Math.log(2)));
        str = str.substring(0, str.indexOf("."));
        Long answer = new Long(str);
        return answer;
    }

    @Test
    public void sampleTests() {
        assertEquals(Long.valueOf(42L), PaperFolder.fold(new Double(384000000)));
        assertEquals(Long.valueOf(0L), PaperFolder.fold(new Double(0.00005)));
    }

    @Test
    public void additionalTests() {
        assertEquals(Long.valueOf(4L), PaperFolder.fold(new Double(0.0016)));
        assertEquals(Long.valueOf(5L), PaperFolder.fold(new Double(0.0032)));
        assertEquals(Long.valueOf(0L), PaperFolder.fold(new Double(0.0000001)));
        assertEquals(Long.valueOf(0L), PaperFolder.fold(new Double(0)));
        assertEquals(null, PaperFolder.fold(new Double(-1)));
    }

    @Test
    public void randomTests() {
        Random rand = new Random();
        for (int i = 0; i < 1000; i++) {
            Double d = new Double(Math.pow(rand.nextDouble(), (double) (rand.nextInt(20) - 10)));
            String message = "Testing " + d;
            assertEquals(message, SolutionTest.solution(d), PaperFolder.fold(d));
        }
    }

    @Test
    public void invalidParameter() {
        assertEquals(null, PaperFolder.fold(new Double("-1")));
    }
}

個人總結：

Very Interesting Kata! 如果厚度爲整型的話直接位運算是否會更快呢？