Sehr einfaches Hangman-Spiel

Folgt dieser Code den Standardkonventionen und ist nicht redundant? Wie kann ich es für andere verständlicher machen?

Es ist eine Übung aus dem Java-Einführungsbuch:

Schreiben Sie ein Hangman-Spiel, das zufällig ein Wort generiert und den Benutzer dazu auffordert, jeweils einen Buchstaben zu erraten. Jeder Buchstabe im Wort wird als Stern angezeigt. Wenn der Benutzer eine korrekte Schätzung vornimmt, wird der tatsächliche Buchstabe angezeigt. Wenn der Benutzer ein Wort beendet hat, zeigen Sie die Anzahl der Fehler an.

Mein Programm:

import java.util.Arrays;
import java.util.Scanner;

public class Hangman{
    public static void main(String[] args) {
        String[] words = {"writer", "that", "program"};
        // Pick random index of words array
        int randomWordNumber = (int) (Math.random() * words.length);
        // Create an array to store already entered letters
        char[] enteredLetters = new char[words[randomWordNumber].length()];
        int triesCount = 0;
        boolean wordIsGuessed = false;
        do {
        // infinitely iterate through cycle as long as enterLetter returns true
        // if enterLetter returns false that means user guessed all the letters
        // in the word e. g. no asterisks were printed by printWord
        switch (enterLetter(words[randomWordNumber], enteredLetters)) {
            case 0:
                triesCount++;
                break;
            case 1:
                triesCount++;
                break;
            case 2:
                break;
            case 3:
                wordIsGuessed = true;
                break;
        }
        } while (! wordIsGuessed);
        System.out.println("\nThe word is " + words[randomWordNumber] +
            " You missed " + (triesCount -findEmptyPosition(enteredLetters)) +
            " time(s)");
    }

    /* Hint user to enter a guess letter,
    returns 0 if letter entered is not in the word (counts as try),
    returns 1 if letter were entered 1st time (counts as try),
    returns 2 if already guessed letter was REentered,
    returns 3 if all letters were guessed */
    public static int enterLetter(String word, char[] enteredLetters)    {
        System.out.print("(Guess) Enter a letter in word ");
        // If-clause is true if no asterisks were printed so
        // word is successfully guessed
        if (! printWord(word, enteredLetters))
            return 3;
        System.out.print(" > ");
        Scanner input = new Scanner(System.in);
        int emptyPosition = findEmptyPosition(enteredLetters);
        char userInput = input.nextLine().charAt(0);
        if (inEnteredLetters(userInput, enteredLetters)) {
            System.out.println(userInput + " is already in the word");
            return 2;
        }
        else if (word.contains(String.valueOf(userInput))) {
            enteredLetters[emptyPosition] = userInput;
            return 1;
        }
        else {
            System.out.println(userInput + " is not in the word");
            return 0;
            }
    }

    /* Print word with asterisks for hidden letters, returns true if
    asterisks were printed, otherwise return false */
    public static boolean printWord(String word, char[] enteredLetters) {
        // Iterate through all letters in word
        boolean asteriskPrinted = false;
        for (int i = 0; i < word.length(); i++) {
            char letter = word.charAt(i);
            // Check if letter already have been entered bu user before
            if (inEnteredLetters(letter, enteredLetters))
                System.out.print(letter); // If yes - print it
            else {
                System.out.print('*');
                asteriskPrinted = true;
            }
        }
        return asteriskPrinted;
    }

    /* Check if letter is in enteredLetters array */
    public static boolean inEnteredLetters(char letter, char[] enteredLetters) {
        return new String(enteredLetters).contains(String.valueOf(letter));
    }

    /* Find first empty position in array of entered letters (one with code \u0000) */
    public static int findEmptyPosition(char[] enteredLetters) {
        int i = 0;
        while (enteredLetters[i] != '\u0000') i++;
        return i;
    }
}

Protokoll seiner Arbeit:

(Guess) Enter a letter in word **** > a
(Guess) Enter a letter in word **a* > t
(Guess) Enter a letter in word t*at > q
q is not in the word
(Guess) Enter a letter in word t*at > t
t is already in the word
(Guess) Enter a letter in word t*at > b
b is not in the word
(Guess) Enter a letter in word t*at > h
(Guess) Enter a letter in word that
The word is that You missed 2 time(s)