Java 刽子手代码

Question

我已经被这段代码困扰了一段时间，并且非常感谢一些帮助。所以基本上我的刽子手代码运行得很好，但是控制台中的显示(其中破折号应该反射(reflect)哪些字母已被正确猜测，哪些字母仍有待猜测)是完全错误的。我将在这里发布我的代码。由于篇幅较长，我只贴出相关方法。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

问题是这一行:

lettersInWord.remove(index);当您调用此方法时，您正在移动数组列表，因此当您获取新索引时，它将是不正确的，因为它已被移动。例如，假设 secret 词是“测试”。首先，数组列表将为 [t,e,s,t]，当您询问第一个 t 的索引时，您将得到 0(正确)。但在调用 arraylist.remove 之后，你将得到 [e.s.t]。当询问第二个 t 的索引时，它将返回 2 而不是您要查找的 3。

也许您应该有第二个数组列表，它可以完整地保存您将用于索引查找的 secret 单词，并保留第一个数组列表来存储剩余的字母。