我的代码会列出一组的所有分区。代码来自此网站：Generating the Partitions of a Set。
我不想将分区打印出来，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，原始网站帖子中的评论仍然存在）：
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的结果如下所示（为了格式化，方括号替换为花括号）：
{{{1,2,3,4}}，$(document).on('beforepopuploads', popupCallback);
，{{2,3,4}}，
$(document).on('beforepopuploads', popCallback);
没有＆＃34;中间＆＃34;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
答案 0 :(得分：1)
我花了一段时间，但我找到了解决方案！我所做的是将所有可能的部分从数组中取出，然后对剩下的部分进行递归，然后添加我作为递归中返回的分区的一部分取出的部分。然后进入一个包含所有可能分区的大数组。为了强制某种顺序我做了它，以便我们取出的这个部分总是采用第一个元素。这样你就不会得到像[[1]，[2,3]]和[[2,3]，[1]]这样的结果，这些结果基本上就是同一个分区。
public static int[][][] getAllPartitions(int[] array) throws Exception {
int[][][] res = new int[0][][];
int n = 1;
for (int i = 0; i < array.length; i++) {
n *= 2;
}
for (int i = 1; i < n; i += 2) {
boolean[] contains = new boolean[array.length];
int length = 0;
int k = i;
for (int j = 0; j < array.length; j++) {
contains[j] = k % 2 == 1;
length += k % 2;
k /= 2;
}
int[] firstPart = new int[length];
int[] secondPart = new int[array.length - length];
int p = 0;
int q = 0;
for (int j = 0; j < array.length; j++) {
if (contains[j]) {
firstPart[p++] = array[j];
} else {
secondPart[q++] = array[j];
}
}
int[][][] partitions;
if (length == array.length) {
partitions = new int[][][] {{firstPart}};
} else {
partitions = getAllPartitions(secondPart);
for (int j = 0; j < partitions.length; j++) {
int[][] partition = new int[partitions[j].length + 1][];
partition[0] = firstPart;
System.arraycopy(partitions[j], 0, partition, 1, partitions[j].length);
partitions[j] = partition;
}
}
int[][][] newRes = new int[res.length + partitions.length][][];
System.arraycopy(res, 0, newRes, 0, res.length);
System.arraycopy(partitions, 0, newRes, res.length, partitions.length);
res = newRes;
}
return res;
}