The challenge
Build a function that returns an array of integers from n to 1 where n>0.
Example : n=5 –> [5,4,3,2,1]
The solution in Java code
Option 1:
public class Sequence{
public static int[] reverse(int n) {
if (n<=0) return null;
int[] out = new int[n];
for (int i=n, j=0; i>0; i--) {
out[j] = i;
j++;
}
return out;
}
}
Option 2:
import java.util.stream.IntStream;
public class Sequence{
public static int[] reverse(int n) {
return IntStream.range(-n, 0).map(Math::abs).toArray();
}
}
Option 3:
import java.util.stream.IntStream;
public class Sequence{
public static int[] reverse(int n) {
return IntStream.iterate(n, i -> i - 1).limit(n).toArray();
}
}
Test cases to validate our solution
import org.junit.Test;
import static org.junit.Assert.assertArrayEquals;
import org.junit.runners.JUnit4;
public class SolutionTest {
private int[] reverseSolution(int n){
int[] result = new int[n];
for(int i=n;i>0;i--){
result[n-i]=i;
}
return result;
}
@Test
public void simpleTest() {
assertArrayEquals(new int[]{5,4,3,2,1},Sequence.reverse(5));
assertArrayEquals(new int[]{6,5,4,3,2,1},Sequence.reverse(6));
assertArrayEquals(reverseSolution(100),Sequence.reverse(100));
assertArrayEquals(reverseSolution(1000),Sequence.reverse(1000));
assertArrayEquals(reverseSolution(100000),Sequence.reverse(100000));
assertArrayEquals(reverseSolution(10000000),Sequence.reverse(10000000));
}
@Test
public void randomTest(){
for(int i=0;i<100;i++){
int random = 1 + (int)(Math.random() * 9999);
assertArrayEquals(reverseSolution(random),Sequence.reverse(random));
}
}
}
