java - 在Java中如何根据单词的长度来定义猜测的次数？

Question

当程序要求输入的次数与单词长度一样多时，我希望获得一些关于该部分编码的帮助。

因此，当我们启动程序时，它会从数组中随机选择一个单词，例如，如果我们要查找的单词是“team”，则意味着我们有四个猜测。

我试图用下面的代码片段来解决这个问题。看起来很简单，但我错过了一些东西:

while(!guess.equals(choosenWord) && letters <= choosenWord.length()) {
                System.out.print("Your guess is: ");
                guess = sc.nextLine();

我需要一些提示的下一部分是，如果我们在字符中有一些匹配项，则应显示找到的字符，并应在其他任何地方显示“-”符号:例如，该词仍然是“team”，我们'正在寻找，我们的猜测是“baby”，而不是“--a-”将被显示。

感谢您的建议。

import java.util.Scanner;


public class FindTheWord {


    public static void main(String[] args) {

        findTheWord();
    }

    public static void findTheWord() {
        Scanner sc = new Scanner(System.in);
        String[] words = {
            "dog", "cat", "house", "love", "friend", "paper", "summer", "chips",
            "number", "file", "program", "lotto", "work", "funny", "database",
            "team", "profile", "facebook", "bean", "winter", "spring", "java", 
            "examination", "hospital", "birth", "baby", "newborn", "airplane",
            "kindergarten", "autumn"};

        int randomWord = (int) (Math.random() * 31);
        String choosenWord = " ";
        int letters = 0;
        String guess = " ";
        char ch = ' ';

        for (int i = 0; i < words[randomWord].length(); i++) {
            choosenWord = words[randomWord];

        }
        for (int j = 0; j < choosenWord.length(); j++) {
            letters = choosenWord.length();


        while(!guess.equals(choosenWord) && letters <= choosenWord.length()) {
            System.out.print("Your guess is: ");
            guess = sc.nextLine();

        }
        }
        System.out.println(words[randomWord]);
        System.out.println(letters);
    }


    }

Answer 1

这是一种使用三元运算符的解决方案 (? :):

...
guess = sc.nextLine();
for (char c : chosenWord.toCharArray()) {
  System.out.print(guess.contains(String.valueOf(c)) ? c : '-');
}
...