无法从代码中删除一行？

Question

该应用程序是一个数字猜谜游戏,我的问题是当用户获得正确的随机数时,它询问用户是否要继续,如果他键入"y"它会创建一个新的随机数并要求用户猜测它.然后它应该询问用户"输入一个数字".相反,我得到了

"太高"(或"太低")
"输入数字"

输出:
输入数字:
2
正确你得到它!数字是2你在2次尝试得到它.
你想再次发挥(Y/N):
Ÿ
太低了!再试一次.
输入数字:

我如何得到它而不是打印"太高"或"太低"的问题,这是在他输入一个数字后确定的？

PS.我尝试了很多方法,但卡住了:(

public static void main(String[] args) {
    System.out
            .println("Welcome to the gussing game, Try to guess the number am thinking of to win!");
    System.out
            .println("++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++");
    System.out.println();

    System.out
            .println("Am thinking of a number between 0 and 10. Try to guess it.");
    System.out.println();

    Scanner sc = new Scanner(System.in);
    String choice = "y";

    double rightNum = Math.random() * 10;
    int randomNum = (int) rightNum; // convert the random number to int
    int tries = 0;

    while (!choice.equalsIgnoreCase("n")) {

        System.out.println("Enter the Number:");
        int guess = sc.nextInt();
        tries++;

        if (guess == randomNum) {
            System.out.println("Correct you've got it! The Number Was "
                    + randomNum);
            System.out.println("You got it in " + tries + " tries.");
            System.out.println("Would you like to play again (y/n):");
            choice = sc.next();

            if (choice.equalsIgnoreCase("y"))
            // reset the random number

            {
                rightNum = Math.random() * 10;
                randomNum = (int) rightNum;
                tries = 0;

            }
        }

        if (guess > randomNum + 10) {
            System.out.println("Way to high! Try again.");
        } else if (guess < randomNum) {
            System.out.println("Too low! Try again.");
        } else if (guess > randomNum && guess <= randomNum + 10) {
            System.out.println("Too high! Try again.");
        }

    }

}

}

Answer 1

else if (guess > randomNum + 10)
{
    System.out.println("Way to high! Try again.");
} 

你错过了其他的.