如何查找列表列表中的哪些项目等于另一个列表

Question

我有一个如下所示的列表：

[[0],
[0, 1, 2],
[2],
[3],
[4],
[5],
[0, 1, 2, 3, 4, 5, 6, 7],
[7],
[8],
[9],
[8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18],
[11],
[11, 12, 13, 14, 15, 16, 17, 18],
[13],
[14],
[14, 15, 16, 17, 18],
[16, 17, 18],
[17],
[17, 18]]

我试图在列表中找到最少数量的项目，当连接时，等于列表的整个范围。在这种情况下，列表的完整范围是这样的：

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18]

所以在这种情况下，列表列表中的这两项将等于整个范围：

[0]
[0, 1, 2]
[2]
[3]
[4]
[5]
---> [0, 1, 2, 3, 4, 5, 6, 7]
[7]
[8]
[9]
---> [8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18]
[11]
[11, 12, 13, 14, 15, 16, 17, 18]
[13]
[14]
[14, 15, 16, 17, 18]
[16, 17, 18]
[17]
[17, 18]

Answer 1

使用itertools.permutations 和chain 的一种方式：

from itertools import permutations, chain

starget = sorted(target)
for i in range(2, len(target)):
    for perm in permutations(l, i):
        if sorted(chain(*perm)) == starget:
            print(i, perm)
            break
    break

输出：

2 ([0, 1, 2, 3, 4, 5, 6, 7], [8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18])