public String transform(int number) {
    boolean isNegative = false;
    StringBuilder sb = new StringBuilder();
    if (number == 0) {
      return "0";
    } else if (number < 0) {
      isNegative = true;
    }
    number = Math.abs(number);

    while (number > 0) {
      sb.append(number % 10);
      number /= 10;
    }

    if (isNegative) {
      sb.append("-");
    }

    return sb.reverse().toString();
  }
Algorithm implementation based on the module of a given number and iterate until the number is
zero. With this approach we get a complexity order equals to O(N) in time terms where N is the
number of digits in the given integer. In space terms, the complexity order is also O(N)
because we are using an StringBuilder as auxiliary data structure.
src/main/java/com/github/pedrovgs/problem11/IntToString.java