0

我有一个 JAVA 任务,我必须Hangman使用数组和循环创建一个程序。

  • user1 输入一个有效的单词(没有数字或符号)
  • 用户 2 可以尝试一次猜出整个单词,或者在总共 10 次尝试中猜出一个字母。最初,用户 2 必须按 1 来猜测单词或按 2 来选择字母。我改变了这一点,因为我认为这对用户更友好。
  • user2 可以在任何时间点尝试猜测单词。

程序需要检查user2的输入是否有效

  • 必须是字母字符,而不是符号
  • 必须只有 1 个字符长度或与要猜测的单词长度相同)。
  • 一个字母字符不能使用两次

如果 user2 的输入无效(上述条件),它会给出错误消息并要求 user2 输入其他内容。任何无效输入不计入 10 次尝试。

目前,如果输入无效(上面的前 2 个条件),则代码的行为应如此。它会给出适当的错误消息,并且尝试次数不会增加。

但是,我似乎无法编写一个条件,如果已经选择了一个字母,它也会给出一条错误消息并要求另一个字母。

我尝试(if upperAlphabet[index] == '*', System.out.println("Duplicate. Try again"))在第一个 do/while 循环中添加一个 if 条件,但它无法正常工作:它增加了尝试次数。

我的印象是我必须在某处做一个 for 循环。找不到地点和方式。

import java.util.Scanner;
import java.util.regex.Pattern;

public class Test {

public static void main(String[] args) {

    char[] upperAlphabet = { 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I',
            'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U',
            'V', 'W', 'X', 'Y', 'Z' };  // Alphabet array to display to user2. 

    String wordtoGuess;
    char letterChoice;
    String userChoiceString;
    String wordArraytoString;

    do {
        System.out.println("Please enter a valid word (letters only)");   // Asks user1 for a valid word
        Scanner wordInput = new Scanner(System.in);
        wordtoGuess = wordInput.next();
        wordtoGuess = wordtoGuess.toUpperCase();
    } while (Pattern.matches("[A-Z]+", wordtoGuess) == false);     // Checks word is valid

    char[] wordArray = wordtoGuess.toCharArray();   // Puts word in character array               
    char[] guessingWordArray = new char[wordtoGuess.length()];
    for (int h = 0; h < guessingWordArray.length; h++) 
        guessingWordArray[h] = '*'; // Displays the word to guess with * for user2

    for (int i = 0; i < 20; i++) {  // Prints 20 empty lines to hide the input of the word from user1 
        System.out.println();
    }

    for (int j = 0; j < 10; j++) {   // 10 attempts loop

        do {

            System.out.print("Word to guess: ");
            System.out.println(guessingWordArray);
            System.out
                    .println("Please choose a letter or solve the word.    " // Asks for a letter or the whole word
                            + "Attempts left: " + (10 - j));
            System.out.println(upperAlphabet);
            Scanner userInput = new Scanner(System.in);
            userChoiceString = userInput.next();
            userChoiceString = userChoiceString.toUpperCase();    // Captures the input as a string
            letterChoice = userChoiceString.charAt(0);
            letterChoice = Character.toUpperCase(letterChoice);  // Captures the first letter of the input

            if (Character.isLetter(letterChoice) == false) // Error if input is an alphabet letter
                System.out.println("Invalid letter. Please try again.");
            if (userChoiceString.length() > 1 // Error if input is not the same length as the whole word but more than 1 character
                    && userChoiceString.length() < wordtoGuess.length())
                System.out.println(("Choose only one letter. Try again."));

        } while (userChoiceString.length() != 1
                && userChoiceString.length() != wordtoGuess.length()
                || Character.isLetter(letterChoice) == false);

        if (userChoiceString.length() == 1) { // if input is only 1 character

            for (int k = 0; k < upperAlphabet.length; k++) { // A used letter is replaced by * in alphabet array.
                if (letterChoice == upperAlphabet[k]) {
                    upperAlphabet[k] = '*';
                }
            }

            for (int m = 0; m < wordtoGuess.length(); m++) { // If a letter is correct, reveal the correct letter in the word to guess.
                if (letterChoice == wordArray[m]) {
                    guessingWordArray[m] = wordArray[m];
                }
            }
            wordArraytoString = new String(guessingWordArray); // If all letters are revealed in the word to guess, display winning message when count of guesses.
            if (wordArraytoString.equals(wordtoGuess)) {

                System.out.println(guessingWordArray);
                System.out.print("Congratulations.");
                System.out.print("You guessed the word: ");
                System.out.print(wordtoGuess);
                System.out.println(" in " + (j + 1) + " guesses.");
                break;

            }

        } else if (userChoiceString.length() == wordtoGuess.length()) { // If user2 tries to guess the whole word, displays winning message and number of guesses
            if (userChoiceString.equals(wordtoGuess)) {
                System.out.println(guessingWordArray);
                System.out.print("Congratulations.");
                System.out.print("You guessed the word: ");
                System.out.print(wordtoGuess);
                if (j == 0)
                    System.out.println(" in " + (j + 1) + " guess.");
                else
                    System.out.println(" in " + (j + 1) + " guesses.");
                break;
            } else
                System.out.println("Wrong guess. Please try again."); // If guessing word is wrong.
        }

        if (j >= 9)
            System.out
                    .println("You did not guess the word in the number of attemps allowed. Better luck next time."); // If exceeds 10 tries.
    }

}

}
4

3 回答 3

1

upperAlphabet当用户进行猜测时,您已经获得了正在修改的数组。也许你可以安排一些事情,如果猜测已经从upperAlphabet,提示用户重复猜测。

你为什么不移动这个循环

for (int k = 0; k < upperAlphabet.length; k++) { // A used letter is replaced by * in alphabet array.
    if (letterChoice == upperAlphabet[k]) {
        upperAlphabet[k] = '*';
    }
}

do/while在提示用户输入的循环中增加几行。确保它只在他们只猜测一个字符时运行。

然后,您可以boolean found = false;在它之前和零件found = true;内部添加该行if。然后紧接着循环,检查 的值found,如果它仍然为假,则显示一条消息,如果用户重复猜测,就会出现这种情况。

do/while如果猜测不是 ,您仍然需要想办法让循环重复found。所以这不是一个完整的答案,但它应该足以让你重新开始。

于 2013-11-07T18:49:24.783 回答
0

因此,为了扩展大卫华莱士的答案,您的代码的这种修改现在可以满足您的需求(尽管它仍然需要他提到的附加条件):

import java.util.Scanner;
import java.util.regex.Pattern;

public class Test {

public static void main(String[] args) {

    char[] upperAlphabet = { 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I',
            'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U',
            'V', 'W', 'X', 'Y', 'Z' };  // Alphabet array to display to user2. 

    int[] guessedLetters = new int[26];

    boolean guessed = false;

    for(int i=0;i<26;i++)
    {
        guessedLetters[i] = 0;
    }

    String wordtoGuess;
    char letterChoice;
    String userChoiceString;
    String wordArraytoString;

    do {
        System.out.println("Please enter a valid word (letters only)");   // Asks user1 for a valid word
        Scanner wordInput = new Scanner(System.in);
        wordtoGuess = wordInput.next();
        wordtoGuess = wordtoGuess.toUpperCase();
    } while (Pattern.matches("[A-Z]+", wordtoGuess) == false);     // Checks word is valid

    char[] wordArray = wordtoGuess.toCharArray();   // Puts word in character array               
    char[] guessingWordArray = new char[wordtoGuess.length()];
    for (int h = 0; h < guessingWordArray.length; h++) 
        guessingWordArray[h] = '*'; // Displays the word to guess with * for user2

    for (int i = 0; i < 20; i++) {  // Prints 20 empty lines to hide the input of the word from user1 
        System.out.println();
    }

    for (int j = 0; j < 10; j++) {   // 10 attempts loop

        do {

            guessed = false;

            System.out.print("Word to guess: ");
            System.out.println(guessingWordArray);
            System.out
                    .println("Please choose a letter or solve the word.    " // Asks for a letter or the whole word
                            + "Attempts left: " + (10 - j));
            System.out.println(upperAlphabet);
            Scanner userInput = new Scanner(System.in);
            userChoiceString = userInput.next();
            userChoiceString = userChoiceString.toUpperCase();    // Captures the input as a string
            letterChoice = userChoiceString.charAt(0);
            letterChoice = Character.toUpperCase(letterChoice);  // Captures the first letter of the input

            if (Character.isLetter(letterChoice) == false) // Error if input is an alphabet letter
                System.out.println("Invalid letter. Please try again.");
            else if (userChoiceString.length() > 1 // Error if input is not the same length as the whole word but more than 1 character
                    && userChoiceString.length() < wordtoGuess.length())
                System.out.println(("Choose only one letter. Try again."));
            for (int k = 0; k < upperAlphabet.length; k++) { // A used letter is replaced by * in alphabet array.
                if(guessedLetters[k] == 1)
                {
                    guessed = true;
                    System.out.println("You've already tried this letter. Please try again.");
                }
                if (letterChoice == upperAlphabet[k]) {
                    //upperAlphabet[k] = '*';
                    guessedLetters[k] = 1; //note which letter has been chosen
                }
            }
        } while (userChoiceString.length() != 1
                && userChoiceString.length() != wordtoGuess.length()
                || Character.isLetter(letterChoice) == false
                || guessed == true);

        if (userChoiceString.length() == 1) { // if input is only 1 character
            /*
            for (int k = 0; k < upperAlphabet.length; k++) { // A used letter is replaced by * in alphabet array.
                if (letterChoice == upperAlphabet[k]) {
                    //upperAlphabet[k] = '*';
                    guessedLetters[k] = 1;
                }
            }
            */

            for (int m = 0; m < wordtoGuess.length(); m++) { // If a letter is correct, reveal the correct letter in the word to guess.
                if (letterChoice == wordArray[m]) {
                    guessingWordArray[m] = wordArray[m];
                }
            }
            wordArraytoString = new String(guessingWordArray); // If all letters are revealed in the word to guess, display winning message when count of guesses.
            if (wordArraytoString.equals(wordtoGuess)) {

                System.out.println(guessingWordArray);
                System.out.print("Congratulations.");
                System.out.print("You guessed the word: ");
                System.out.print(wordtoGuess);
                System.out.println(" in " + (j + 1) + " guesses.");
                break;

            }

        } else if (userChoiceString.length() == wordtoGuess.length()) { // If user2 tries to guess the whole word, displays winning message and number of guesses
            if (userChoiceString.equals(wordtoGuess)) {
                System.out.println(guessingWordArray);
                System.out.print("Congratulations.");
                System.out.print("You guessed the word: ");
                System.out.print(wordtoGuess);
                if (j == 0)
                    System.out.println(" in " + (j + 1) + " guess.");
                else
                    System.out.println(" in " + (j + 1) + " guesses.");
                break;
            } else
                System.out.println("Wrong guess. Please try again."); // If guessing word is wrong.
        }

        if (j >= 9)
            System.out
                    .println("You did not guess the word in the number of attemps allowed. Better luck next time."); // If exceeds 10 tries.
    }

}

}
于 2013-11-07T20:04:23.593 回答
0

您可能想要做的只是创建另一个数组来将您看到的字符推入而不是覆盖 upperAlphabet 中的值,这样发现它更容易且不那么混乱。

ArrayList<char> guessed = new ArrayList();
....
do{
  ...
while(//your conditions here
     && !guessed.contains(letterChoice) );
guessed.add(letterChoice));

您也可以使用常规数组执行此操作,如下所示。

char guessed = new char[10];
do{
  ...
while(//your conditions here
      &&!contains(guessed, letterChoice));
add(guessed, letterChoice)

//supporting methods
public boolean contains(char[] arr, char val){
  //check that array has value
}

public void add(char[] arr, char val){
  //add val to first empty space in arr.
}
于 2013-11-07T18:49:42.030 回答