Java刽子手代码

Question

我已经被这段代码困扰了一段时间，非常感谢您的帮助。因此，基本上我的刽子手代码运行良好，但控制台中的显示，破折号应该反映哪些字母猜对了，哪些字母仍然需要猜对，这是完全错误的。我将在这里发布我的代码。由于时间很长，我将只发布相关的方法。我觉得问题似乎出在显示数组列表上，我无法持续更新它。

     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

问题在于这一行：

lettersInWord.remove（index）；当你调用这个时，你正在移动数组列表，所以当你得到新的索引时，它将是不正确的，因为它已经被移动了。例如，让我们说秘密词是“测试”。起初，数组列表将是[t， e， s， t]，当你要求第一个t的索引时，你会得到0（正确）。但是打电话arraylist.remove你会有[e. s. t]。当请求第二个t的索引时，它将返回2而不是您要查找的3。

也许你应该有一个第二个数组列表，它保持秘密单词不变，你将使用它进行索引查找，并保持第一个数组列表来存储剩余的字母。