Java 石头剪刀布游戏 [关闭]

我是编程新手，我正在尝试用 Java 编写一个非常简单的石头剪刀布游戏。它将编译并运行良好，但我希望说一些类似“无效的移动。再试一次”之类的内容。或当用户（personPlay）未输入正确的字符（r、p 或 s）时类似的内容。最好的方法是什么？例如，如果您输入“q”，它应该打印“无效移动”。提前非常感谢！

// *************
// Rock.java 
// ************* 

import java.util.Scanner; 
import java.util.Random; 


public class Rock 
{ 
public static void main(String[] args) 
{ 
    String personPlay; //User's play -- "R", "P", or "S" 
    String computerPlay = ""; //Computer's play -- "R", "P", or "S" 
    int computerInt; //Randomly generated number used to determine 
                     //computer's play 
    String response; 


    Scanner scan = new Scanner(System.in); 
    Random generator = new Random(); 

    System.out.println("Hey, let's play Rock, Paper, Scissors!\n" + 
                       "Please enter a move.\n" + "Rock = R, Paper" + 
                       "= P, and Scissors = S.");

    System.out.println();

    //Generate computer's play (0,1,2) 
    computerInt = generator.nextInt(3)+1; 

    //Translate computer's randomly generated play to 
    //string using if //statements 

    if (computerInt == 1) 
       computerPlay = "R"; 
    else if (computerInt == 2) 
       computerPlay = "P"; 
    else if (computerInt == 3) 
       computerPlay = "S"; 


    //Get player's play from input-- note that this is 
    // stored as a string 
    System.out.println("Enter your play: "); 
    personPlay = scan.next();

    //Make player's play uppercase for ease of comparison 
    personPlay = personPlay.toUpperCase(); 

    //Print computer's play 
    System.out.println("Computer play is: " + computerPlay); 


    //See who won. Use nested ifs 

    if (personPlay.equals(computerPlay)) 
       System.out.println("It's a tie!"); 
    else if (personPlay.equals("R")) 
       if (computerPlay.equals("S")) 
          System.out.println("Rock crushes scissors. You win!!");
    else if (computerPlay.equals("P")) 
            System.out.println("Paper eats rock. You lose!!"); 
    else if (personPlay.equals("P")) 
       if (computerPlay.equals("S")) 
       System.out.println("Scissor cuts paper. You lose!!"); 
    else if (computerPlay.equals("R")) 
            System.out.println("Paper eats rock. You win!!"); 
    else if (personPlay.equals("S")) 
         if (computerPlay.equals("P")) 
         System.out.println("Scissor cuts paper. You win!!"); 
    else if (computerPlay.equals("R")) 
            System.out.println("Rock breaks scissors. You lose!!"); 
    else 
         System.out.println("Invalid user input."); 
}

}

我建议制作石头、剪刀、布的物体。这些对象将具有与字符串之间进行转换的逻辑，并且还“知道”什么击败什么。 Java 枚举非常适合此目的。

public enum Type{

  ROCK, PAPER, SCISSOR;

  public static Type parseType(String value){
     //if /else logic here to return either ROCK, PAPER or SCISSOR

     //if value is not either, you can return null
  }
}

The parseType方法可以返回null如果 String 不是有效类型。您的代码可以检查该值是否为空，如果是，则打印“无效重试”并循环返回以重新读取扫描仪。

Type person=null;

 while(person==null){
      System.out.println("Enter your play: "); 
      person= Type.parseType(scan.next());
      if(person ==null){
         System.out.println("invalid try again");
      }
 }

此外，您的类型枚举可以通过让每个类型来确定什么击败什么Type对象知道：

public enum Type{

    //...

    //each type will implement this method differently
    public abstract boolean beats(Type other);


}

每种类型都会以不同的方式实现此方法，看看哪种方法更胜一筹：

ROCK{

   @Override
   public boolean beats(Type other){            
        return other ==  SCISSOR;

   }
}

 ...

然后在你的代码中

 Type person, computer;
   if (person.equals(computer)) 
   System.out.println("It's a tie!");
  }else if(person.beats(computer)){
     System.out.println(person+ " beats " + computer + "You win!!"); 
  }else{
     System.out.println(computer + " beats " + person+ "You lose!!");
  }