Java Hangman Code

Question

我已经坚持使用这段代码一段时间了，非常感谢一些帮助。所以基本上我的hangman代码运行正常，但是控制台中的显示，其中破折号应该反映哪些字母已被正确猜测，以及哪些字母仍有待猜测是完全错误的。我会在这里发布我的代码。由于时间很长，我只会发布相关方法。我觉得问题就在于显示器阵列列表，我无法持续更新。

     public void play() {

     String SecretWord = getWord();
     ArrayList <String> lettersInWord = new ArrayList <String>();
     for (int i=0;i<SecretWord.length();i++){
            lettersInWord.add(Character.toString(SecretWord.charAt(i)));
            }
     int remainingChances = getNumberGuesses();
     int noOfLetters = SecretWord.length(); 
     ArrayList<String> lowerCaseAlphabets = getAlphabetArrayList();
     ArrayList<String> display = new ArrayList<String>();
     for (int i = 0; i<noOfLetters; i++) {
         display.add("_");
     }
     System.out.println(printDisplay(display));
     Set <String> lettersGuessed = new HashSet<String>();
     while (remainingChances > 0){
         String question = readString("What letter do you want to guess?");
         if (lowerCaseAlphabets.contains(question)){
         System.out.println("Number of misses remaining equals "+remainingChances+"");
         int index = lettersInWord.indexOf(question);
            if (index== -1){
                lettersGuessed.add(question);
                remainingChances-= 1;
                if (remainingChances==0){
                    System.out.println("No "+question+". You lose! The secret word was "+SecretWord+"");
                }
                else if (remainingChances>0){
                    System.out.println("There is no "+question+" in the word");
                    System.out.println(printDisplay(display));
                    System.out.println("Guesses so Far :"+lettersGuessed+"");

                }
            }
            else if (index!=-1){
                while (index!= -1){
                    display.set(index, question);
                    System.out.println(printDisplay(display));
                    System.out.println("Guesses so Far :"+lettersGuessed+"");
                    lettersInWord.remove(index);
                    if (lettersInWord.size()==0){
                        System.out.println("You have won! Congratulations!");
                        return;
                    }
                    else if (lettersInWord.size()!=0){
                        index = lettersInWord.indexOf(question);

                    }
                }

            }
         }
         else {
             System.out.println("The letter you have chosen in invalid. You must pick a lower case letter from the alphabet!");

         }
     }
}            

    public String printDisplay(ArrayList<String> display){
    String View = "";
    for (int i =0;i<display.size();i++){
        View+= display.get(i) + " ";
    }
    return View;
}   

Answer 1

问题是这一行：

lettersInWord.remove(index); 当你调用它时，你正在移动arraylist，这样当你获得新索引时，它将是不正确的，因为它已被移动。例如，让我们说秘密词是“测试”。首先，arraylist将是[t，e，s，t]，当你要求第一个t的索引时，你将得到0（正确）。但是在调用arraylist.remove之后你会有[e.s.t]。当询问第二个t的索引时，它将返回2而不是你要找的3个。

也许你应该有一个第二个arraylist，保持秘密单词完整，你将用于索引查找，并保留第一个存储其余字母。