简单的Java方法？

Question

好的，所以我需要在java中创建一个基本上是随机数生成器的程序，玩家需要猜测从1-10生成的数字。

我已经获得了代码，一切都运行得很好（代码发布在下面），但我想这样做，以便用户不能使用相同的数字两次。

比如说，例如玩家再次输入2然后是3然后是2。我想要它 返回＆＃34;您已输入此号码＆＃34;对于我的生活，我无法让它发挥作用。

我一直在尝试使用HashSets，因为我相当确定该方法是正确的方法，但是当我回想起HashSet中已有的数字时，我会被搞砸他们已经在HashSet中然后返回＆＃34;您已经输入了这个号码＆＃34;。

我试着修补

if (hs.contains(guess)) == true) 
System.out.println("You have already entered this number")

我也放hs.add(guess)并猜测玩家输入了什么号码。我真的很难过，如果你们能指导我朝着正确的方向前进，我会非常感激。 非常感谢先进的

package guessgame;

import java.util.HashSet;
import java.util.Scanner;

public class GuessGame {

    public static void main(String[] args) {

        HashSet<Integer>hs = new HashSet();
        int GuessLogic;
        GuessLogic = (int) (Math.random() * 10 + 1);
        Scanner keyboard = new Scanner (System.in);
        int guess;
        int NumGuess;
        NumGuess = 1;

        do {
            System.out.print("Enter a guess: ");
            guess = keyboard.nextInt();
            System.out.println("Your guess is " + guess);
            if (guess == GuessLogic)
                System.out.println("You got it right!! Congrats!! Total Number of Guesses: " + NumGuess++);
            else if (guess < GuessLogic && guess > 0)
                System.out.println("You are wrong!!! Hint: Guess Higher, Guess number: " + NumGuess++);
            else if (guess > GuessLogic && guess <= 10)
                System.out.println("You are wrong!!! Hint: Guess Lower, Guess number: " + NumGuess++); 
            else
                System.out.println("Your guess is out of the specified range. Please try again." ); 
        } while (guess != GuessLogic);

    }   

} 

Answer 1

首先，你应该编程到界面 Set（而不是具体的类型HashSet）。接下来，如果此集合尚未包含指定的元素，则可以使用Set.add(E)返回 true。最后，我建议使用有意义的变量名。将它们放在一起，可能看起来像，

Set<Integer> hs = new HashSet<>();
int correctNumber = 1 + (int) (Math.random() * 10);
Scanner keyboard = new Scanner(System.in);
int guessCount = 1;
int guess;
do {
    System.out.print("Enter a guess: ");
    guess = keyboard.nextInt();
    System.out.println("Your guess is " + guess);
    if (hs.add(guess)) {
        if (guess == correctNumber) {
            System.out.println("You got it right!! Congrats!! Total Number of "
                    + "Guesses: " + guessCount++);
        } else if (guess < correctNumber && guess > 0) {
            System.out.println("You are wrong!!! Hint: Guess "
                    + "Higher, Guess number: " + guessCount++);
        } else if (guess > correctNumber && guess <= 10) {
            System.out.println("You are wrong!!! Hint: Guess "
                    + "Lower, Guess number: " + guessCount++);
        } else {
            System.out.println("Your guess is out of the specified range. "
                    + "Please try again.");
        }
    } else {
        System.out.println("You have already entered this number");
    }
} while (guess != correctNumber);

Answer 2

使用HashSet。如果add函数已包含该对象，则返回false：

Set<Integer> guesses = new HashSet<Integer>();
if (guesses.add(guess)) {
    //Code for their new guess
} else {
    System.out.println("You have already guessed this number!");
}

Answer 3

这是完整版（重写了一点）。

您可能希望将NumGuess++移动到想要增加猜测次数的位置。在猜测之后立即放置它似乎是合适的，但如果Set已包含该数字，您可能不想增加。

此外，移出范围检查可使代码更清晰。

public static void main(String[] args) {

    Scanner keyboard = new Scanner(System.in);

    HashSet<Integer> hs = new HashSet<>();
    int GuessLogic = (int) (Math.random() * 10 + 1);

    int guess;
    int NumGuess = 0;
    do {
        System.out.print("Enter a guess: ");
        guess = keyboard.nextInt();
        NumGuess++;

        if (hs.contains(guess)) {
            System.out.println("You have already entered this number");
            continue; // this will repeat the loop
        }

        if (guess < 0 || guess > 10) {
            System.out.println("Your guess is out of the specified range. Please try again.");
            continue; // this will repeat the loop
        }

        System.out.println("Your guess is " + guess);
        if (guess == GuessLogic) {
            System.out.println("You got it right!! Congrats!! Total Number of Guesses: " + NumGuess);
            return; // this will stop the method
        }
        else if (guess < GuessLogic) {
            System.out.println("You are wrong!!! Hint: Guess Higher, Guess number: " + NumGuess);
        }
        else if (guess > GuessLogic) {
            System.out.println("You are wrong!!! Hint: Guess Lower, Guess number: " + NumGuess);
        }

        hs.add(guess);

    } while (true);

}

Answer 4

我使用以下代码：

package guessgame;

import java.util.ArrayList;
import java.util.Scanner;

public class GuessGame {
    public static void main(String[] args) {
        ArrayList<Integer> guesses = new ArrayList<>(); // Arraylist preferrable to HashSet, but not required
        int GuessLogic;
        GuessLogic = (int)(Math.random() * 10 + 1);
        Scanner keyboard = new Scanner(System.in);
        int guess;
        int NumGuess;
        NumGuess = 1;
        do {
            System.out.print("Enter a guess: ");
            guess = keyboard.nextInt();
            if (guesses.contains(guess)) { // If the player has already guessed the number
                System.out.println("You have already entered this number");
                continue; // Skips to the next loop iteration
            }
            guesses.add(guess); // ONLY does this if ^ evaluates to false
            System.out.println("Your guess is " + guess);
            if (guess == GuessLogic)
                System.out.println("You got it right!! Congrats!! Total Number of Guesses: " + NumGuess++);
            else if (guess < GuessLogic && guess > 0)
                System.out.println("You are wrong!!! Hint: Guess Higher, Guess number: " + NumGuess++);
            else if (guess > GuessLogic && guess <= 10)
                System.out.println("You are wrong!!! Hint: Guess Lower, Guess number: " + NumGuess++); 
            else
                System.out.println("Your guess is out of the specified range. Please try again." ); 
        } while (guess != GuessLogic);
    }    
}

Answer 5

对于如此小范围的值，您只需使用boolean数组就可以逃脱：

int minRngVal = 1;
int maxRngVal = 10;
boolean[] guesses = new boolean[maxRngVal];

当玩家猜到一个数字时，将其发送到如下方法：

private boolean isValidGuess(int guess){
    if(guesses[guess-1] == true){
        //player already guessed that number!
        return false;
    } else{
        //player hasn't guessed that number yet
        guesses[guess-1] = true;
        return false;
    }
}

然后只需在代码中添加以下行：

if (isValidGuess(guess)) {
    System.out.println("You have already entered this number");
    continue; // this will repeat the loop
}