java - 刽子手错误的猜测无法正确打印

Question

我正在为学校开发这个 Hangman 程序，当用户猜错字母时，我无法正确打印错误猜测的数量。这是我到目前为止得到的代码，我将不胜感激任何提示。

import java.util.Scanner;

public class HangmanTest {
    public static void main(String[] args) {

        String[] wordBank = { "madelynn", "crystal", "mcbride", "daughter",
                "adorable", "beautiful", "andrew", "programming", "alyssa",
                "computers", "mcbreezy", "maddy", "happy", "vacation", "beach",
                "java", "benefical", "military", "veteran", "standale",
                "lions", "tigers", "redwings", "pistons", "michigan",
                "football", "baseball", "hockey", "basketball", "golf" };
        int minimum = 0;
        int maximum = wordBank.length - 1;
        String again;

        do {
            int choice = minimum + (int) (Math.random() * maximum);

            String word = wordBank[choice];

            // Converts the random word to asterix
            String userWord = "";
            for (int i = 0; i < word.length(); i++) {
                userWord += "*";
            }

            // Breaks into a bunch of characters
            char[] userWordCh = userWord.toCharArray();

            // Show the random word
            System.out.println("The word for you to guess is " + userWord);

            // instantiate a scanner object
            Scanner input = new Scanner(System.in);

            int size = word.length();
            int rightGuesses = 0;
            int wrongGuesses = 0;

            while (size != rightGuesses) {
                System.out.println("Enter a character: ");
                String response = input.next();
                char ch = response.charAt(0);

                char[] wordChars = word.toCharArray();

                for (int i = 0; i < word.length(); i++) {
                    if (wordChars[i] == ch) {
                        userWordCh[i] = ch;
                        ++rightGuesses;
                    } else {
                        ++wrongGuesses;
                    }
                } // end of for loop

                System.out.print("The word is: ");
                for (int j = 0; j < userWordCh.length; j++)
                    System.out.print(userWordCh[j]);

                System.out.println();
            } // end of while loop

            System.out.println("You had " + wrongGuesses + " wrong guesses.");

            System.out.println("Would you like to play again y/n: ");
            again = input.next();

        } while (again.equals("y"));

    }
}

Answer 1

for (int i = 0; i < word.length(); i++) {
    if (wordChars[i] == ch) {
        userWordCh[i] = ch;
        ++rightGuesses;
    } else {
        ++wrongGuesses;
    }
} // end of for loop

在此循环中，每次猜测与单词中的字母匹配时，rightGuesses 都会增加 1，而每次猜测与单词中的字母不匹配时，rightGuesses 都会增加 1。正如您可以想象的那样，这将导致数字总体上增加与字母数量相同的数字，而它应该只增加一次。

尝试如下:

boolean foundMatch = false;
for (int i = 0; i < word.length(); i++) {
    if (wordChars[i] == ch) {
        userWordCh[i] = ch;
        if (!foundMatch)
        {
            ++rightGuesses;
            foundMatch = true;
        }
    }
}
if (!foundMatch)
{
    ++wrongGuesses;
}
// end of for loop

现在我们只增加 rightGuesses 和rongGuesses 之一一次 - rightGuesses 仅在我们没有找到匹配项时才可以增加(将找到的匹配设置为 true)，而 falseGuesses 只能在我们没有找到匹配项时增加一次。