Octonumero (ver 2)

An octonumero of a number x is found as follows.

1. Convert x to base 8, call this p
2. Sort the digits of p in increasing order, call this q
3. Subtract q from p in octal (convert q and p to decimal, subtract and convert back to octal)
4. Repeat steps 2 - 3, four more times or until the digits in the result are in sorted order.
5. Convert the number back to decimal

Example:

x = 3418₁₀

convert: 3418₁₀ = 6532₈

sort and subtract: 6532₈ - 2356₈ = 4154₈

4 more times: 4154₈ - 1445₈ = 2507₈

3 more times: 2507₈ - 257₈ = 2230,

2 more times: 2230₈ - 223₈ = 2005₈

last time: 2005₈ - 25₈ = 1760₈

convert: 1760₈ = 1008₁₀

I need help with the following methods, please:

1. public static int[] numToArray (int num)

//Precondition: 999 < num < 10000. Assume num is a 4 digit number

numToArraywill place every digit of num as an element of an array

For example, int [] digits = numToArray(2534)will result in an array calleddigits that contains{ 2, 5, 3, 4}

1. public static int[] sortArray (int[] unsorted)

sortArraywill return an array in ascending order.

For example, if digits contains { 2, 5, 3, 4 } a call to int[] sorted = sorted(digits)will result in an array calledsortedthat contains

{ 2, 3, 4, 5 }.