Question

通过for循环循环时，该循环的长度为1-10。我也经历了一个while循环。我试图找到1-42之间的6个随机数。然后我再做一次（1-10）次。目前，这6个随机数保持不变。我怀疑它与种子有关，但我无法解决。

我尝试同时使用SecureRandom，Random和通过Math.random（）什么都没有。

模型：

    public static final int MAX_TIPS = 10;
    public static final int MIN_TIPS = 1;
    public static final int REQUIRED_NUMBERS = 6;
    public static final int RANGE_NUMBER_MIN = 1;
    public static final int RANGE_NUMBER_MAX = 42;
    public static final int RANGE_LUCKY_NUMBER_MIN = 1;
    public static final int RANGE_LUCKY_NUMBER_MAX = 6;

    private Random random = new Random();

    private int getRandomInt(int min, int max) {
         return random.nextInt((max - min) + 1) + min;
    }

    public ArrayList<LottoTip> createOpponentTips() {
        ArrayList<Integer> opponentChosenNumbers = new ArrayList<>();
        ArrayList<LottoTip> lottoTips = new ArrayList<>();

        //1-10
        int amountOfTips = getRandomInt(MIN_TIPS, MAX_TIPS);

        for (int i = 0; i < amountOfTips; i++) {

            int[] arrayOpponentChosenNumbers = new int[REQUIRED_NUMBERS];

            while (opponentChosenNumbers.size() < REQUIRED_NUMBERS) {

                //1-42 This here is the problem zone

                int randomNumber = getRandomInt(RANGE_NUMBER_MIN, RANGE_NUMBER_MAX);
                if (!opponentChosenNumbers.contains(randomNumber)) {
                    opponentChosenNumbers.add(randomNumber);
                }
            }
            //1-6 ---This here works
            int opponentLuckyNumber = getRandomInt(RANGE_LUCKY_NUMBER_MIN, RANGE_LUCKY_NUMBER_MAX);

            for (int j = 0; j < opponentChosenNumbers.size(); j++) {
                arrayOpponentChosenNumbers[j] = opponentChosenNumbers.get(j);
            }

            lottoTips.add(new LottoTip(arrayOpponentChosenNumbers, opponentLuckyNumber));
        }

        return lottoTips;
    }

LottoTip：

public class LottoTip {
    private int[] numbers;
    private int luckyNumber;

    public LottoTip(int[] numbers, int luckyNumber) {
        this.numbers = numbers;
        this.luckyNumber = luckyNumber;
    }

    public String getNumbers() {
        return numbers[0] + ", " + numbers[1] + ", " + numbers[2] + ", " + numbers[3] + ", " + numbers[4] + ", " + numbers[5];
    }

    public void setNumbers(int[] numbers) {
        this.numbers = numbers;
    }

    public int[] getNumbersArray() {
        return numbers;
    }

    public int getLuckyNumber() {
        return luckyNumber;
    }

    public void setLuckyNumber(int luckyNumber) {
        this.luckyNumber = luckyNumber;
    }

    @Override
    public String toString() {
        return "\nNumbers: " + getNumbers() + " Lucky Number: " + luckyNumber;
    }
}

这是一个System.out.println，因为您可以看到数字保持不变。您能帮我找到一种创建可以一直工作的getRandomInt方法的方法吗？

非常感谢您的帮助。

Kerry York [
Numbers: 6, 4, 30, 15, 25, 5 Lucky Number: 2, 
Numbers: 6, 4, 30, 15, 25, 5 Lucky Number: 4, 
Numbers: 6, 4, 30, 15, 25, 5 Lucky Number: 5, 
Numbers: 6, 4, 30, 15, 25, 5 Lucky Number: 1, 
Numbers: 6, 4, 30, 15, 25, 5 Lucky Number: 3, 
Numbers: 6, 4, 30, 15, 25, 5 Lucky Number: 6]

Don Dickinson [
Numbers: 23, 29, 34, 18, 16, 19 Lucky Number: 6, 
Numbers: 23, 29, 34, 18, 16, 19 Lucky Number: 5, 
Numbers: 23, 29, 34, 18, 16, 19 Lucky Number: 3, 
Numbers: 23, 29, 34, 18, 16, 19 Lucky Number: 2, 
Numbers: 23, 29, 34, 18, 16, 19 Lucky Number: 4, 
Numbers: 23, 29, 34, 18, 16, 19 Lucky Number: 1, 
Numbers: 23, 29, 34, 18, 16, 19 Lucky Number: 1, 
Numbers: 23, 29, 34, 18, 16, 19 Lucky Number: 4, 
Numbers: 23, 29, 34, 18, 16, 19 Lucky Number: 3]

Clifford Weinstein [
Numbers: 6, 33, 13, 2, 37, 38 Lucky Number: 6, 
Numbers: 6, 33, 13, 2, 37, 38 Lucky Number: 3, 
Numbers: 6, 33, 13, 2, 37, 38 Lucky Number: 2, 
Numbers: 6, 33, 13, 2, 37, 38 Lucky Number: 1, 
Numbers: 6, 33, 13, 2, 37, 38 Lucky Number: 4, 
Numbers: 6, 33, 13, 2, 37, 38 Lucky Number: 3, 
Numbers: 6, 33, 13, 2, 37, 38 Lucky Number: 5, 
Numbers: 6, 33, 13, 2, 37, 38 Lucky Number: 6, 
Numbers: 6, 33, 13, 2, 37, 38 Lucky Number: 3, 
Numbers: 6, 33, 13, 2, 37, 38 Lucky Number: 3]

Angela Spencer [
Numbers: 12, 39, 8, 25, 40, 15 Lucky Number: 6, 
Numbers: 12, 39, 8, 25, 40, 15 Lucky Number: 3, 
Numbers: 12, 39, 8, 25, 40, 15 Lucky Number: 2, 
Numbers: 12, 39, 8, 25, 40, 15 Lucky Number: 4, 
Numbers: 12, 39, 8, 25, 40, 15 Lucky Number: 5, 
Numbers: 12, 39, 8, 25, 40, 15 Lucky Number: 4, 
Numbers: 12, 39, 8, 25, 40, 15 Lucky Number: 2, 
Numbers: 12, 39, 8, 25, 40, 15 Lucky Number: 2, 
Numbers: 12, 39, 8, 25, 40, 15 Lucky Number: 3]

Leroy McIntosh [
Numbers: 5, 23, 2, 24, 28, 3 Lucky Number: 1]

Answer 1

您可以在外部for循环的迭代之间重用opponentChosenNumbers列表。

这样，在第二次迭代中，while循环后卫

while (opponentChosenNumbers.size() < REQUIRED_NUMBERS) {

立即为假，因此您不会选择新的随机数。

将opponentChosenNumbers的声明移至for循环中（或确保在while循环之前为空，例如，通过调用opponentChosenNumbers.clear()）。

随机生成器创建相同的数字

1 个答案: