查找具有元素 1...N 的列表的 K 个子集，同时保留元素的顺序

Question

给定一个来自 1...N 的整数列表，我试图找到元素的 K 个子集，同时保留元素的顺序。例如，当 N = 4 且 K = 2 时：

[1] [2, 3, 4]

[1, 2] [3, 4]

[1, 2, 3] [4]

[1, 2, 3, 4] []

应该是正确的输出。 到目前为止，我已经得到了第一列可能性。但我很难得到正确的逻辑。

    final = [['' for x in range(K)] for y in range(N)]
    i = 0
    for k in range(0, K):
        # row tracker
        i = 0
        while i < N:
            if k > 0:
                st = len(final[i][k - 1])
            else:
                st = 0
            for j in range(0, N):
                tmp = ""
                prefix = chemicals[:j + 1]
                tmp = tmp.join(str(i) for i in prefix)
                final[i][k] = tmp
                i += 1
        print

同样，正确的输出应该是：

[1] [2, 3, 4]

[1, 2] [3, 4]

[1, 2, 3] [4]

[1, 2, 3, 4] []

集合可以是空的。

更新：这是 N=4、K=3 的正确输出

[1] [2] [3, 4]
[1] [2, 3] [4]
[1] [2, 3, 4] []
[1, 2] [3] [4]
[1, 2] [3, 4] []
[1, 2, 3] [4] []
[1, 2, 3, 4] [] []

Answer 1

我认为一个简单的带有切片的列表理解就足够了。您可能还想使用itertools.combinations：

import itertools

N = 4
K = 2

elements = list(range(1, N + 1))
final = [[elements[a:b] for a, b in zip([0] + cuts, cuts + [N])]
                        for cuts in (list(c) for c in itertools.combinations(elements, K - 1))]

for x in final:
    print(*x)

输出：

[1] [2, 3, 4]
[1, 2] [3, 4]
[1, 2, 3] [4]
[1, 2, 3, 4] []

Answer 2

您可以使用从给定起始编号迭代索引的函数，默认为1，到n，产生从起始编号到索引的数字范围，并递归连接来自递归的子集使用一个高一个子集和一个小一个子集的起始索引进行调用，直到起始索引大于n 或k 变为1，此时应产生剩余范围：

def get_subsets(n, k, s=1):
    if s > n or k == 1:
        yield [list(range(s, n + 1))] + [[] for _ in range(1, k)]
        return
    for i in range(s, n + 1):
        for subsets in get_subsets(n, k - 1, i + 1):
            yield [list(range(s, i + 1))] + subsets

这样：

for s in get_subsets(4, 2):
    print(*s)

输出：

[1] [2, 3, 4]
[1, 2] [3, 4]
[1, 2, 3] [4]
[1, 2, 3, 4] []

还有：

for s in get_subsets(4, 3):
    print(*s)

输出：

[1] [2] [3, 4]
[1] [2, 3] [4]
[1] [2, 3, 4] []
[1, 2] [3] [4]
[1, 2] [3, 4] []
[1, 2, 3] [4] []
[1, 2, 3, 4] [] []