java - 返回原始开始位置，无需退出应用程序并再次打开它

Question

该应用程序的目的非常简单。要求用户输入一个号码，并检查用户号码和随机创建的号码是否相同。直到用户输入正确的数字为止。一旦用户输入正确的数字，我如何更改我的随机数，以便他们可以继续而不退出应用程序。现在，一旦找到正确的数字，我就必须退出。然后再次打开它。
公共(public)类 MainActivity 扩展 AppCompatActivity { 整数；//全局变量。

    public void clicked (View view ){
       EditText userinput = (EditText)findViewById(R.id.userinput);
       String inputstring = userinput.getText().toString();
       if (inputstring == " "){
          Toast.makeText(getApplicationContext(),"You havent entered anything",Toast.LENGTH_SHORT).show();
    }

    int inputint = Integer.parseInt(inputstring);

    if (inputint > number ){
        Toast.makeText(getApplicationContext(),"No, too high",Toast.LENGTH_SHORT).show();
    }
    else if (inputint < number){
        Toast.makeText(getApplicationContext(),"No, too small",Toast.LENGTH_SHORT).show();
    }
    else{
        Toast.makeText(getApplicationContext(),"Well Done! You Guessed it",Toast.LENGTH_SHORT).show();

    }

        //System.out.println("Computer guessed number is: " + number);
}
@Override
protected void onCreate(Bundle savedInstanceState) {
    Random r = new Random();
    number = r.nextInt(21);
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
}

}

Answer 1

你可以尝试这样的随机函数

public void randomGenerator()
{
  Random r = new random
  number = r.nextInt(21) //put the int number creation in global
}

然后在onCreate()中调用它

    @Override
protected void onCreate(Bundle savedInstanceState) {

    randomGenerator()
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
}

也在 else 语句中

else{
        Toast.makeText(getApplicationContext(),"Well Done! You Guessed it",Toast.LENGTH_SHORT).show();

        randomGenerator(); //random number after user get correct answer

    }