我有一个 JAVA 作业,必须使用数组和循环创建一个 Hangman
程序。
- 用户1输入一个有效的单词(无数字或符号)
- 用户2可以尝试一次猜出整个单词,或者使用一个字母总共猜10次。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。
- 用户2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
如果用户2的输入无效(上述条件),它会给出错误消息并要求用户2输入其他内容。任何无效输入都不会计入 10 次尝试。
目前,如果输入无效(上述前 2 个条件),代码将按其应有的方式运行。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,它还会给出错误消息并要求输入另一个字母。
我尝试在第一个 do/while 中放入 if 条件 (if upperAlphabet[index] == '*', System.out.println("Duplicate. Try Again"))
循环,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个地方做一个 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.
}
}
}
最佳答案
您已经获得了数组 upperAlphabet
,当用户进行猜测时您正在修改该数组。也许您可以安排一些事情,以便如果猜测超出了 upperAlphabet
,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
循环中增加几行,提示用户输入。确保它仅在他们只猜出一个字符时运行。
然后,您可以在其前面添加行 booleanfound = false;
,并在 if
部分内添加 found = true;
。然后紧接着循环,检查 found
的值,如果仍然为 false,则显示一条消息(如果用户重复猜测,就会出现这种情况)。
如果没有找到
猜测,您仍然需要找到一种方法来重复执行do/while
循环。所以这不是一个完整的答案,但它应该足以让你再次前进。
关于java - Hangman 赋值和循环条件,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/19843392/