我有一个JAVA作业，必须使用数组和循环创建一个Hangman程序。
user1输入一个有效的单词（没有数字或符号）
user2可以一口气猜出整个单词，也可以尝试10次，用一个字母猜出一个单词。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，因为我认为这更加用户友好。
user2可以在任何时间尝试猜测单词。
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
只能为1个字符长度或与要猜测的单词相同的长度）。
字母字符不能两次使用
如果user2的输入无效（上述条件），它将给出错误消息，并要求user2输入其他内容。任何无效输入均不会计入10次尝试。
现在，如果输入无效（上面的前两个条件），则代码的行为应与预期相同。它会给出适当的错误消息，并且尝试次数不会增加。
但是，我似乎无法编写这样的条件：如果已经选择了一个字母，那么它还会给出错误消息并要求另一个字母。
我尝试在第一个do / while循环中放入if条件(if upperAlphabet[index] == '*', System.out.println("Duplicate. Try again"))，但它无法正常工作：它会增加尝试次数。
我的印象是我必须在某个地方进行for循环。无法找到位置和方式。
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.
}
}
}
最佳答案
您已经有了要在用户猜测时修改的数组upperAlphabet。也许您可以安排一些事情，以便如果猜测已从upperAlphabet中删除，则系统会提示用户重复猜测。
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] = '*';
}
}
在
do/while循环中几行提示用户输入。确保仅在他们猜一个字符的情况下运行它。然后，您可以在行之前添加
boolean found = false;，并在found = true;部分内添加if。然后，紧随循环之后，检查found的值，并显示一条消息，如果它仍然仍然为false，则用户重复猜测就会出现这种情况。如果猜测不是
do/while，您仍然需要找到一种方法使found循环重复。因此，这不是一个完整的答案，但足以让您重新出发。