You have an array of numbers.
Your task is to sort ascending odd numbers but even numbers must be on their places.
Zero isn't an odd number and you don't need to move it. If you have an empty array, you need to return it.
Example
sortArray([5, 3, 2, 8, 1, 4]) == [1, 3, 2, 8, 5, 4]
Good Solution1:
import java.util.*;
public class Kata {
public static int[] sortArray(final int[] array) {
// Sort the odd numbers only
final int[] sortedOdd = Arrays.stream(array).filter(e -> e % 2 == 1).sorted().toArray();
// Then merge them back into original array
for (int j = 0, s = 0; j < array.length; j++) {
if (array[j] % 2 == 1) array[j] = sortedOdd[s++];
}
return array;
}
}
Good Solution2:
import java.util.PrimitiveIterator.OfInt;
import java.util.stream.IntStream;
public class Kata {
public static int[] sortArray(int[] array) {
OfInt sortedOdds = IntStream
.of(array)
.filter(i -> i % 2 == 1)
.sorted()
.iterator();
return IntStream
.of(array)
.map(i -> i % 2 == 0 ? i : sortedOdds.nextInt())
.toArray();
}
}
