給定來自 1...NI 的整數列表，我試圖找到元素的 K 個子集，同時保留元素的順序。例如，當 N = 4 且 K = 2 時：[1] [2, 3, 4][1, 2] [3, 4][1, 2, 3] [4][1, 2, 3, 4] []將是正確的輸出。到目前為止，我已經獲得了第一列的可能性。但我正在努力獲得正確的邏輯。 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同樣，正確的輸出將是：[1] [2, 3, 4][1, 2] [3, 4][1, 2, 3] [4][1, 2, 3, 4] []集合可以為空的地方。更新：這是 N=4, K=3 的正確輸出[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] [] []
2 回答
慕妹3242003
TA貢獻1824條經驗 獲得超6個贊
您可以使用一個函數從給定的起始編號迭代索引，默認為1, to n，產生從起始編號到索引的一系列數字，并遞歸地將遞歸調用中的子集與一個更高的起始索引和少一個子集，直到起始索引大于n或k變為1，此時應產生剩余范圍：
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
以便：
for s in get_subsets(4, 2):
print(*s)
輸出：
[1] [2, 3, 4]
[1, 2] [3, 4]
[1, 2, 3] [4]
[1, 2, 3, 4] []
然后：
for s in get_subsets(4, 3):
print(*s)
輸出：
[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] [] []
子衿沉夜
TA貢獻1828條經驗 獲得超3個贊
我認為對切片進行簡單的列表理解就足夠了。您可能還想使用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)
輸出：
[1] [2, 3, 4]
[1, 2] [3, 4]
[1, 2, 3] [4]
[1, 2, 3, 4] []
添加回答
舉報
0/150
提交
取消