Java中的递归数字排列

Question

最近我被分配了一个这样的作业：

编写一个名为NumberPermutation.java的程序并实现下面指定的递归方法：

public static void permutation(int num)

此方法有一个正整数作为其参数，并显示该长度的奇数位的所有排列。

按升序显示这些值。例如，如果长度为 3，您的程序应显示：

111
113
115
117
119
131
133
135
...

我对 Java 很陌生（过去一年我之前的所有编码任务都是在 Python 中完成的，这是我从事的第三个 Java 程序），所以我完全不知道从哪里获得开始了。如果有人能帮助我并告诉我如何开始，我将不胜感激，谢谢。

Answer 1

虽然我不喜欢“帮我做作业”的问题，但这次我会提供答案。这是我的建议。请不要只是复制粘贴上交。先了解一下，通过这个例子了解一些Java的基础知识。

public class RecursionTester {
    public static void main(String[] args) {
        new RecursionTester().permutation(3);
    }

    private void permutation(int num) {
        permutation(num, "");
    }

    private void permutation(int num, String buffer) {
        if (num == 0) {
            System.out.println(buffer);
            return;
        }
        for (int i = 1; i < 10; i += 2) {
            permutation(num - 1, buffer + Integer.toString(i));
        }
    }
}

结果只是转到 System.out 的事实使这个解决方案毫无用处，但如果任务只是证明您了解递归，那应该就足够了。

Answer 2

map 和 reduce 方法

您可以将此任务解释为获取n-th 个奇数数字 数组的笛卡尔积。如果是n=3，那么你会得到5<sup>3</sup>=125 组合。

mapToObj 方法准备二维数组进行求和并指向结果的格式：

1: {{1}, {3}, {5}, {7}, {9}}
2: {{1}, {3}, {5}, {7}, {9}}
3: {{1}, {3}, {5}, {7}, {9}}

reduce 方法将 2D 数组对相加成一个 2D 数组 - 如果 n=3 减少了两个步骤：

1: {{1}, {3}, {5}, {7}, {9}}
2: {{1}, {3}, {5}, {7}, {9}}
---
sum1: {{1, 1}, {1, 3}, {1, 5}, ...}
3: {{1}, {3}, {5}, {7}, {9}}
---
total: {1, 1, 1}, {1, 1, 3}, ...}

Try it online!

public static int[][] cartesianProduct(int n, int[][] arr) {
    return IntStream.range(0, n)
            // stream of n-th number of 2D arrays
            .mapToObj(i -> arr)
            // stream of 2D arrays into a single 2D array
            .reduce((arr1, arr2) -> Arrays.stream(arr1)
                // combinations of rows of two arrays
                .flatMap(row1 -> Arrays.stream(arr2)
                    // concatenate into a single row
                    .map(row2 -> Stream.of(row1, row2)
                        .flatMapToInt(Arrays::stream)
                        // row of a 2D array
                        .toArray()))
                // 2D array of combinations
                .toArray(int[][]::new))
            // otherwise an empty array
            .orElse(new int[0][]);
}

public static void main(String[] args) {
    // exponent
    int n = 3;
    // 2D array of odd digits
    int[][] arr = {{1}, {3}, {5}, {7}, {9}};
    // cartesian product
    int[][] cp = cartesianProduct(n, arr);
    // output
    System.out.println("Number of combinations: " + cp.length);
    // number of rows
    int rows = cp.length / arr.length;
    // column-wise output
    IntStream.range(0, rows)
            .mapToObj(i -> IntStream.range(0, cp.length)
                .filter(j -> j % rows == i)
                .mapToObj(j -> Arrays.toString(cp[j]))
                .collect(Collectors.joining(" ")))
            .forEach(System.out::println);
}

输出：

Number of combinations: 125
[1, 1, 1] [3, 1, 1] [5, 1, 1] [7, 1, 1] [9, 1, 1]
[1, 1, 3] [3, 1, 3] [5, 1, 3] [7, 1, 3] [9, 1, 3]
[1, 1, 5] [3, 1, 5] [5, 1, 5] [7, 1, 5] [9, 1, 5]
[1, 1, 7] [3, 1, 7] [5, 1, 7] [7, 1, 7] [9, 1, 7]
[1, 1, 9] [3, 1, 9] [5, 1, 9] [7, 1, 9] [9, 1, 9]
[1, 3, 1] [3, 3, 1] [5, 3, 1] [7, 3, 1] [9, 3, 1]
[1, 3, 3] [3, 3, 3] [5, 3, 3] [7, 3, 3] [9, 3, 3]
[1, 3, 5] [3, 3, 5] [5, 3, 5] [7, 3, 5] [9, 3, 5]
[1, 3, 7] [3, 3, 7] [5, 3, 7] [7, 3, 7] [9, 3, 7]
[1, 3, 9] [3, 3, 9] [5, 3, 9] [7, 3, 9] [9, 3, 9]
[1, 5, 1] [3, 5, 1] [5, 5, 1] [7, 5, 1] [9, 5, 1]
[1, 5, 3] [3, 5, 3] [5, 5, 3] [7, 5, 3] [9, 5, 3]
[1, 5, 5] [3, 5, 5] [5, 5, 5] [7, 5, 5] [9, 5, 5]
[1, 5, 7] [3, 5, 7] [5, 5, 7] [7, 5, 7] [9, 5, 7]
[1, 5, 9] [3, 5, 9] [5, 5, 9] [7, 5, 9] [9, 5, 9]
[1, 7, 1] [3, 7, 1] [5, 7, 1] [7, 7, 1] [9, 7, 1]
[1, 7, 3] [3, 7, 3] [5, 7, 3] [7, 7, 3] [9, 7, 3]
[1, 7, 5] [3, 7, 5] [5, 7, 5] [7, 7, 5] [9, 7, 5]
[1, 7, 7] [3, 7, 7] [5, 7, 7] [7, 7, 7] [9, 7, 7]
[1, 7, 9] [3, 7, 9] [5, 7, 9] [7, 7, 9] [9, 7, 9]
[1, 9, 1] [3, 9, 1] [5, 9, 1] [7, 9, 1] [9, 9, 1]
[1, 9, 3] [3, 9, 3] [5, 9, 3] [7, 9, 3] [9, 9, 3]
[1, 9, 5] [3, 9, 5] [5, 9, 5] [7, 9, 5] [9, 9, 5]
[1, 9, 7] [3, 9, 7] [5, 9, 7] [7, 9, 7] [9, 9, 7]
[1, 9, 9] [3, 9, 9] [5, 9, 9] [7, 9, 9] [9, 9, 9]

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^{另见：How do I generate a Cartesian product in Java?}