集合的分区 - 存储导致一系列嵌套列表

我有代码将列出一组的所有分区 . 代码来自此站点：Generating the Partitions of a Set .

我不想只打印出分区，而是将它们存储为列表 . 我想在这个递归示例中返回的内容之后对我的结果进行建模：How to find all partitions of a set .

我想要一个整数列表的列表 . 内部列表是中间列表中包含的分区的子集，外部列表包含所有分区的完整集合 .

这是我的代码（从C转换为Java，从原始网站发布的评论仍然存在）：

import java.util.ArrayList;
import java.util.List;


public class PartitionApp {


  public static class PNR {
    static
    /*
    	printp
    		- print out the partitioning scheme s of n elements 
    		as: {1, 2, 4} {3}
    */
    ArrayList < ArrayList < ArrayList < Integer >>> outerList = new ArrayList < > ();
    public static void PNR(int[] s, int n) {
      /* Get the total number of partitions. In the example above, 2.*/

      int part_num = 1;
      int i;

      for (i = 0; i < n; ++i)
        if (s[i] > part_num) {

          part_num = s[i];
        }
        /* Print the p partitions. */

      int p;

      for (p = part_num; p >= 1; --p) {

        System.out.print("{");
        ArrayList < Integer > innerList = new ArrayList < > ();
        ArrayList < ArrayList < Integer >> middleList = new ArrayList < > ();
        /* If s[i] == p, then i + 1 is part of the pth partition. */
        for (i = 0; i < n; ++i) {
          if (s[i] == p) {
            innerList.add(i + 1);
            System.out.print(i + 1);
            System.out.print(",");
          }

        }
        middleList.add(innerList);
        outerList.add(middleList);

        System.out.print("} ");
      }

      System.out.print("\n");
      System.out.println(outerList);

    }

    /*
	next
		- given the partitioning scheme represented by s and m, generate
		the next

	Returns: 1, if a valid partitioning was found
		0, otherwise
*/
    static int next(int[] s, int[] m, int n) {
      /* Update s: 1 1 1 1 -> 2 1 1 1 -> 1 2 1 1 -> 2 2 1 1 -> 3 2 1 1 ->
	1 1 2 1 ... */
      /*int j;
	printf(" -> (");
	for (j = 0; j < n; ++j)
		printf("%d, ", s[j]);
	printf("\b\b)\n");*/
      int i = 0;
      ++s[i];
      while ((i < n - 1) && (s[i] > m[i + 1] + 1)) {
        s[i] = 1;
        ++i;
        ++s[i];
      }

      /* If i is has reached n-1 th element, then the last unique partitiong
	has been found*/
      if (i == n - 1)
        return 0;

      /* Because all the first i elements are now 1, s[i] (i + 1 th element)
	is the largest. So we update max by copying it to all the first i
	positions in m.*/
      if (s[i] > m[i])
        m[i] = s[i];
      for (int j = i - 1; j >= 0; --j) {
        m[j] = m[i];

      }


      /*	for (i = 0; i < n; ++i)
      		printf("%d ", m[i]);
      	getchar();*/
      return 1;
    }

    public static void main(String[] args) {
      int count = 0;
      int[] s = new int[16];
      /* s[i] is the number of the set in which the ith element
      			should go */
      int[] m = new int[16]; /* m[i] is the largest of the first i elements in s*/

      int n = 4;
      int i;
      /* The first way to partition a set is to put all the elements in the same
	   subset. */
      for (i = 0; i < n; ++i) {
        s[i] = 1;
        m[i] = 1;
      }

      /* Print the first partitioning. */
      PNR(s, n);

      /* Print the other partitioning schemes. */
      while (next(s, m, n) != 0) {
        PNR(s, n);
        count++;
      }
      count = count + 1;
      System.out.println("count = " + count);


      //	return 0;
    }

  }

}

我得到的n = 4的结果看起来像这样（方括号被大括号替换为格式化目的）：

{{{1,2,3,4}}，{{1}}，{{2,3,4}}，{{2}}，{{1,3,4}}，{{1,2 }，{{3,4}} .....

没有“中间”分组 . 所有内部子集（应该是一组n个元素的一部分）都包含在外部集合中的列表中 . 我没有正确设置内部，中间和外部列表，并且一直在努力解决这个问题 . 我希望有人可以帮我看看我的错误 .

谢谢你，丽贝卡