Very much agreed with @Patrik M, but the thing with Arrays.toString is that it includes "[" and "]" and "," in the output. So I'll simply use a regex to remove them from outout like this

String strOfInts = Arrays.toString(intArray).replaceAll("\\[|\\]|,|\\s", "");

and now you have a String which can be parsed back to java.lang.Number, for example,

long veryLongNumber = Long.parseLong(intStr);

Or you can use the java 8 streams, if you hate regex,

String strOfInts = Arrays
            .stream(intArray)
            .mapToObj(String::valueOf)
            .reduce((a, b) -> a.concat(",").concat(b))
            .get();

Use a Stream which is available from Java 8. To get a Stream instance with "list" of ints:

• For int[]
  • IntStream intStream = Arrays.Stream(nums); or
  • Stream<Integer> intStream = Arrays.Stream(nums).boxed(); if you need the same class as bottom one.
• For any classes with Collection<Integer> interface (ex. Vector<Integer>, List<Integer>)
  • Stream<Integer> intStream = nums.stream();

Finally, to get a String[]:

String[] answer = intStream.sorted().mapToObj(String::valueOf).toArray(String[]::new);

Using Java 8's stream library, we can make this a one-liner (albeit a long line):

String str = "[1, 2, 3, 4, 5, 6, 7, 8, 9, 0]";
int[] arr = Arrays.stream(str.substring(1, str.length()-1).split(","))
    .map(String::trim).mapToInt(Integer::parseInt).toArray();
System.out.println(Arrays.toString(arr));

substring removes the brackets, split separates the array elements, trim removes any whitespace around the number, parseInt parses each number, and we dump the result in an array. I've included trim to make this the inverse of Arrays.toString(int[]), but this will also parse strings without whitespace, as in the question. If you only needed to parse strings from Arrays.toString, you could omit trim and use split(", ") (note the space).