Question

我正在制作乘法表（从2到9） - 这是10个随机生成的样本，例如

2 * 3 = 
4 * 5 = 
... (7 more times)
9 * 5 =

要点是所有样品必须不同并且样品

5 * 8 =

和

8 * 5 =

被认为是相同的

我的想法是创建描述数字对的类对，覆盖它的等于方法，生成随机数，创建Pair并将Pair对象添加到Set。

public class Pair {
    private int first;
    private int second;

    public int getFirst() {
        return first;
    }

    public int getSecond() {
        return second;
    }

    public void setFirst(int first) {
        this.first = first;
    }

    public void setSecond(int second) {
        this.second = second;
    }

    public Pair() {}

    public Pair(int first, int second) {
        this.first = first;
        this.second = second;
    }

    @Override
    public boolean equals(Object o) {
        if (o == null || o.getClass() != this.getClass())
            return false;

        Pair that = (Pair) o;
        return (this.first == that.first && this.second == that.second) ||
                (this.second == that.first && this.first == that.second);
    }

    @Override
    public int hashCode() {
        int result = 17;
        int prime = 31;
        result = result * prime + first;
        result = result * prime + second;
        return result;
    }

    @Override
    public String toString() {
        return first + " * " + second + " =";
    }
}

public static Pair[] createTable(int count) {
        Random random = new Random();
        Set<Pair> set = new HashSet<Pair>();
        while (set.size() < count) {
            int first = random.nextInt(8) + 2;
            int second = random.nextInt(8) + 2;
            set.add(new Pair(first, second));
        }
        return set.toArray(new Pair[count]);
}

问题是方法createTable（）返回的某些数组包含等价对例如在

中

[7 * 6 =, 5 * 6 =, 4 * 8 =, 4 * 9 =, 2 * 8 =, 9 * 2 =, 8 * 2 =, 6 * 3 =, 5 * 2 =, 4 * 2 =]

有2 * 8和8 * 2对不应该在那里

哪里出错？

Answer 1

您的hashCode()方法错误。 hashCode() 必须为两个相等的对象返回相等的值。根据您的equals()方法，对5-7等于对7-5，但它们的hashCode()不一样。

要正确实施hashCode()，您应该始终以最低的数字开头：

public int hashCode() {
    int result = 17;
    int prime = 31;

    int lowest = first < second ? first : second;
    int highest = first < second ? second : first;

    result = result * prime + lowest;
    result = result * prime + highest;
    return result;
}

或更简单：

public int hashCode() {
    int lowest = first < second ? first : second;
    int highest = first < second ? second : first;
    return Objects.hash(lowest, highest);
}

或者，甚至更简单（感谢@ user2336315）：

public int hashCode() {
    return Objects.hash(Math.min(first, second), Math.max(first, second));
}

乘法表

1 个答案: