Question

所以，我成功地制作了一个非常简单但完整的刽子手版本。现在我想改进程序并慢慢添加功能（最终我想要的图形）。我想要修复的第一件事就是重复字母，此时，如果用户决定键入相同的字母（单词中的一个字母）与单词的长度一样多，他们就会得到虚假的“胜利”。什么是阻止这种情况发生的最好方法？

PS。我是编程的新手，这个项目对我来说真的很难......所以如果答案真的很明显，对不起要问，我今晚再也想不到了。

任何帮助将不胜感激，这是我的代码：

import java.io.*;

public class hangman_test2 
{

    public static void main(String[] args) throws IOException 
    {
        BufferedReader in;
        in = new BufferedReader (new InputStreamReader (System.in));

        boolean Lets_play=true;
        String response;

        while (Lets_play)
        {
            printheader();

            String the_word=getWord();

            System.out.println(the_word);
            print_blanks(the_word);     

            guesses(the_word);

            System.out.println("Want to play again?");
            response=in.readLine();
            if(response.charAt(0)=='n' || response.charAt(0)=='N')
            {
                Lets_play= false;
                System.out.println("Thanks for playing!");
            }
        }
    }//end main


    public static void printheader()
    {
    System.out.println("Welcome, lets play hangman!");
    System.out.println("enter letters to guess the word\n");
    }//end print header

    public static String getWord()
    {

        String [] possible_word=new String [10];


        possible_word[0]="green";
        possible_word[1]="orange";
        possible_word[2]="tree";
        possible_word[3]="flowers";
        possible_word[4]="ocean";
        possible_word[5]="grudge";
        possible_word[6]="scraple";
        possible_word[7]="crab";
        possible_word[8]="insect";
        possible_word[9]="stripes";


        String theWord= possible_word [(int)(Math.random()*possible_word.length)];
        return theWord;
    }//end the word

    public static void print_blanks(String the_word)
    {
        for (int x=0; x<the_word.length(); x++)
        {
            System.out.print("_ ");
        }

    }//print blanks

    public static void guesses(String the_word)throws IOException
    {
        BufferedReader in;
        in = new BufferedReader (new InputStreamReader (System.in));

        boolean thisRound=true;
        int strike=0;


        int right_letter=0;


        while (thisRound)
        {

            int letters_not_in_word=0;

            char letter_guessed=in.readLine().charAt(0);


            for (int current_letter=0; current_letter<the_word.length(); current_letter++)
            {

                if (the_word.charAt(current_letter)==letter_guessed)


                {

                    System.out.println(letter_guessed + " fits in space number " + (current_letter+1));

                    right_letter++;


                    if(right_letter == the_word.length())
                    {

                        win(the_word);
                        thisRound=false;
                    }
                }

                else
                {
                    letters_not_in_word++;

                    if (letters_not_in_word==the_word.length())
                    {
                        System.out.println(letter_guessed + " is not in the word");
                        strike ++;

                        if(strike==5)
                        {
                            lose(the_word);
                            thisRound=false;
                        }

                    }//if

                }//else
            }//for

        }//while

    }//end guesses


    public static void win( String word)
    {
    System.out.println("\ncongradulations, you won!");
    System.out.println("the word is " + word + "\n");
    }

    public static void lose( String word)
    {
    System.out.println("\nsorry, you lost");
    System.out.println("the word is " + word + "\n");
    }

}

Answer 1

查看HashSet和Set界面。 Set的作用是防止两次添加同一个对象。

因此，每次用户添加一个字母时，请检查它是否已在该集合中。如果不是，那么检查他们的猜测并将其添加到集合中。如果它已经在集合中，那么忽略他们的猜测。

刽子手游戏中的错误

1 个答案: