傳回1到n的所有組合python_如何用Python列出N個數字的所有排列組合?

itertools.permutations(iterable[, r])

Return successive r length permutations of elements in the iterable.

If r is not specified or is None, then r defaults to the length of the itera

ble and all possible full-length permutations are generated.

Permutations are emitted in lexicographic sort order. So, if the input itera

ble is sorted, the permutation tuples will be produced in sorted order.

Elements are treated as unique based on their position, not on their value.

So if the input elements are unique, there will be no repeat values in each

permutation.

Roughly equivalent to:

def permutations(iterable, r=None):

# permutations('ABCD', 2) --> AB AC AD BA BC BD CA CB CD DA DB DC

# permutations(range(3)) --> 012 021 102 120 201 210

pool = tuple(iterable)

n = len(pool)

r = n if r is None else r

if r > n:

return

indices = range(n)

cycles = range(n, n-r, -1)

yield tuple(pool[i] for i in indices[:r])

while n:

for i in reversed(range(r)):

cycles[i] -= 1

if cycles[i] == 0:

indices[i:] = indices[i+1:] + indices[i:i+1]

cycles[i] = n - i

else:

j = cycles[i]

indices[i], indices[-j] = indices[-j], indices[i]

yield tuple(pool[i] for i in indices[:r])

break

else:

return

【 在 sjtudent (Me) 的大作中提到: 】

: 多謝大俠!!! 不知用代碼可否實作?