How to find the permutation of a sort in Java

Question

I want to sort an array and find the index of each element in the sorted order. So for instance if I run this on the array:

[3,2,4]

I\'d ge

Answer 1

As an update, this is relatively easy to do in Java 8 using the streams API.

public static int[] sortedPermutation(final int[] items) {
  return IntStream.range(0, items.length)
    .mapToObj(value -> Integer.valueOf(value))
    .sorted((i1, i2) -> Integer.compare(items[i1], items[i2]))
    .mapToInt(value -> value.intValue())
    .toArray();
}

It somewhat unfortunately requires a boxing and unboxing step for the indices, as there is no .sorted(IntComparator) method on IntStream, or even an IntComparator functional interface for that matter.

To generalize to a List of Comparable objects is pretty straightforward:

public static > int[] sortedPermutation(final List items) {
  return IntStream.range(0, items.size())
    .mapToObj(value -> Integer.valueOf(value))
    .sorted((i1, i2) -> items.get(i1).compareTo(items.get(i2)))
    .mapToInt(value -> value.intValue())
    .toArray();
}