无阵列的java刽子手游戏

这段代码的目标是创建一个基本的刽子手游戏。直到最后一步，一切正常。在用户猜出所有字母后，它应该显示

“您猜到了”

，但循环似乎额外执行了一次。我尝试将循环代码更改为：

while（猜字母

但这只会让循环过早结束一次。
非常感谢您的帮助

import java.util.Scanner;

public class SecretPhrase {
public static void main(String[] args) {
    char userChoice;
    String secretPhrase = "GO TEAM";
    Scanner input = new Scanner(System.in);
    String hint = "G* T***";
    StringBuilder secretWord = new StringBuilder(hint);

    System.out.println("The hint is " + hint); 

    System.out.println("Please guess a letter");
    userChoice = Character.toUpperCase(input.nextLine().charAt(0)); //Taking the input and turning to uppercase for comparing purposes

    int asteriskAmount = hint.length() - 1; // "-1" for the space.
    int guessedLetters = 2; // for the 'G' and the 'T' that are already displayed.

    while(guessedLetters < asteriskAmount) {
        boolean isInPhrase = checkLetter(userChoice, secretPhrase);
        if(isInPhrase) { // if the guessed letter is a letter in the phrase...
            int position = getPosition(userChoice, secretPhrase); 
            secretWord.setCharAt(position, userChoice);
            hint = secretWord.toString();
            guessedLetters++; 
            System.out.println(secretWord);
            System.out.println("Please guess a letter");
            userChoice = Character.toUpperCase(input.nextLine().charAt(0));
        }
        else {
            System.out.println("That letter is not in the phrase. Please try again >>> ");
            userChoice = Character.toUpperCase(input.nextLine().charAt(0));
        }
    }
    System.out.println("You got it! The secret word is " + secretWord);


}
public static boolean checkLetter(char userChoice, String secretPhrase) {
    boolean isInPhrase = false;
    int amountOfLetters = 0;
    for(int x = 0; x < secretPhrase.length(); x++) {
        if(userChoice != secretPhrase.charAt(x)) {
            isInPhrase = false;
        }
        else {
            isInPhrase = true;
            amountOfLetters++;
        }
    }
    if(amountOfLetters >= 1) {
        isInPhrase = true;
    }
    return isInPhrase;
}
public static int getPosition(char userChoice, String secretPhrase) {
    int position;
    int x = 0;
    while(userChoice != secretPhrase.charAt(x)) {
        x++;
    }
    position = x;
    return position;
}
}

---

问题是，您总是在递增

猜测字母后再次询问。所以它总是会多问一次。尝试重新排序，例如：

//System.out.println("Please guess a letter");
//userChoice = Character.toUpperCase(input.nextLine().charAt(0)); //Taking the input and turning to uppercase for comparing purposes

final int asteriskAmount = hint.length() - 1; // "-1" for the space.
int guessedLetters = 2; // for the 'G' and the 'T' that are already displayed.

while (guessedLetters < asteriskAmount) {
    System.out.println(secretWord);
    System.out.println("Please guess a letter");
    userChoice = Character.toUpperCase(input.nextLine().charAt(0));

    final boolean isInPhrase = checkLetter(userChoice, secretPhrase);
    if (isInPhrase) { // if the guessed letter is a letter in the phrase...
        final int position = getPosition(userChoice, secretPhrase);
        secretWord.setCharAt(position, userChoice);
        hint = secretWord.toString();
        guessedLetters++;
    } else {
        System.out.println("That letter is not in the phrase. Please try again >>> ");
        userChoice = Character.toUpperCase(input.nextLine().charAt(0));
    }

    System.out.println(guessedLetters);
}

//System.out.println（“请猜一个字母”）；
//userChoice=Character.toUpperCase（input.nextLine（）.charAt（0））//获取输入并转换为大写以进行比较
final int asteriskAmount=hint.length（）-1；//“1”表示空间。
int guessedLetters=2；//对于已显示的“G”和“T”。
while（猜字母>”；
userChoice=Character.toUpperCase（input.nextLine（）.charAt（0））；
}
System.out.println（猜字母）；
}
在while循环中获取用户输入。在最后一个循环中，您验证了用户输入，然后在循环结束时请求另一个用户输入。游戏尝试并完成，但在完成之前，它必须在
while
中完成最后一次用户输入。我总是尝试将我的用户输入放在循环的开始，这样类似的事情就不会发生。希望这有帮助

System.out.println("The hint is " + hint); 

//Move the 2 lines below inside your loop
//System.out.println("Please guess a letter");
//userChoice = Character.toUpperCase(input.nextLine().charAt(0)); //Taking the input and turning to uppercase for comparing purposes

int asteriskAmount = hint.length() - 1; // "-1" for the space.
int guessedLetters = 2; // for the 'G' and the 'T' that are already displayed.

while(guessedLetters < asteriskAmount) {

    //Moved from above
    System.out.println(secretWord);
    System.out.println("Please guess a letter");
    userChoice = Character.toUpperCase(input.nextLine().charAt(0));

    boolean isInPhrase = checkLetter(userChoice, secretPhrase);
    if(isInPhrase) { // if the guessed letter is a letter in the phrase...
        int position = getPosition(userChoice, secretPhrase); 
        secretWord.setCharAt(position, userChoice);
        hint = secretWord.toString();
        guessedLetters++; 

        //These are not needed any more
        //System.out.println(secretWord);
        //System.out.println("Please guess a letter");
        //userChoice = Character.toUpperCase(input.nextLine().charAt(0));
    }

System.out.println（“提示为”+hint）；
//在循环内移动下面的两行
//System.out.println（“请猜一个字母”）；
//userChoice=Character.toUpperCase（input.nextLine（）.charAt（0））//获取输入并转换为大写以进行比较
int asteriskAmount=hint.length（）-1；//“1”表示空间。
int guessedLetters=2；//对于已显示的“G”和“T”。
while（猜字母
从循环之前删除这些行

System.out.println("Please guess a letter");
userChoice = Character.toUpperCase(input.nextLine().charAt(0));

现在在循环内部，将这些线移动到循环的开始处

while(guessedLetters < asteriskAmount) {
        System.out.println("Please guess a letter");
        userChoice = Character.toUpperCase(input.nextLine().charAt(0));
    boolean isInPhrase = checkLetter(userChoice, secretPhrase);
    if(isInPhrase) { // if the guessed letter is a letter in the phrase...
        int position = getPosition(userChoice, secretPhrase); 
        secretWord.setCharAt(position, userChoice);
        hint = secretWord.toString();
        guessedLetters++; 
        System.out.println(secretWord);
    }
    else {
        System.out.println("That letter is not in the phrase. Please try again >>> ");
        userChoice = Character.toUpperCase(input.nextLine().charAt(0));
    }
}

嘿，谢谢你的回答。如果没有数组的情况下出现了不止一次，我怎么用字母替换星号呢？我在想，与其使用
checkLetter
方法返回
布尔值，不如使用
int
值来告诉我该字母出现的时间，以及然后做一些循环，用字母替换星号。还有其他建议吗？很抱歉，我没有早点回复，我正在路上。我已经更新了我的答案，在不使用数组的情况下实现了相同的结果。看一看，看看你的想法：）非常感谢你的帮助。我只是想指出，你需要添加中的de>（）
s

啊，我很困惑，因为数组也有

.length

但没有

（）

while(guessedLetters < asteriskAmount) {
        System.out.println("Please guess a letter");
        userChoice = Character.toUpperCase(input.nextLine().charAt(0));
    boolean isInPhrase = checkLetter(userChoice, secretPhrase);
    if(isInPhrase) { // if the guessed letter is a letter in the phrase...
        int position = getPosition(userChoice, secretPhrase); 
        secretWord.setCharAt(position, userChoice);
        hint = secretWord.toString();
        guessedLetters++; 
        System.out.println(secretWord);
    }
    else {
        System.out.println("That letter is not in the phrase. Please try again >>> ");
        userChoice = Character.toUpperCase(input.nextLine().charAt(0));
    }
}

for(int r=0; r<secretPhrase.length(); r++){
    if(secretPhrase.charAt(r) == userChoice){
        secretWord.setCharAt(r, userChoice);
        guessedLetters++;
    }
}

while(guessedLetters < asteriskAmount) {
        System.out.println("Please guess a letter");
        userChoice = Character.toUpperCase(input.nextLine().charAt(0));
    boolean isInPhrase = checkLetter(userChoice, secretPhrase);
    if(isInPhrase) { // if the guessed letter is a letter in the phrase...
        for(int r=0; r<secretPhrase.length(); r++){
            if(secretPhrase.charAt(r) == userChoice){
                secretWord.setCharAt(r, userChoice);
                guessedLetters++;
            }
        }
        System.out.println(secretWord);
    }
    else {
        System.out.println("That letter is not in the phrase. Please try again >>> ");
        userChoice = Character.toUpperCase(input.nextLine().charAt(0));
    }
}