作者bigpigbigpig (To littlepig with love)
看板Python
标题[范例] next_permutation
时间Thu May 29 10:57:31 2014
def next_permutation(L):
n = len(L)
if n == 0 or n == 1: return (False, L)
i = n - 1
while True:
i -= 1
if L[i] < L[i+1]:
j = n - 1
while not ( L[i] < L[j] ): j = j - 1
L[i], L[j] = L[j], L[i]
nL = L[(i+1):]
nL.reverse()
L[(i+1):] = nL
return (True, L)
if i == 0:
L.reverse()
return (False, L)
执行范例：
>>> L1 = [ 1, 2, 3, 4 ]
>>> count = 0
>>> not_yet = True
>>> while not_yet:
count += 1
print("%5d" % count, L1)
(not_yet, L1) = next_permutation(L1)
1 [1, 2, 3, 4]
2 [1, 2, 4, 3]
3 [1, 3, 2, 4]
4 [1, 3, 4, 2]
5 [1, 4, 2, 3]
6 [1, 4, 3, 2]
7 [2, 1, 3, 4]
8 [2, 1, 4, 3]
9 [2, 3, 1, 4]
10 [2, 3, 4, 1]
11 [2, 4, 1, 3]
12 [2, 4, 3, 1]
13 [3, 1, 2, 4]
14 [3, 1, 4, 2]
15 [3, 2, 1, 4]
16 [3, 2, 4, 1]
17 [3, 4, 1, 2]
18 [3, 4, 2, 1]
19 [4, 1, 2, 3]
20 [4, 1, 3, 2]
21 [4, 2, 1, 3]
22 [4, 2, 3, 1]
23 [4, 3, 1, 2]
24 [4, 3, 2, 1]
>>>
--
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※ 编辑: bigpigbigpig (1.169.131.239), 05/29/2014 12:47:30
1F：推 tiefblau:itertool表示 05/29 13:04
2F：→ tjjh89017:这是要实作离散数学中的作法吧XD 05/29 17:32
3F：推 shaopin:so..? C++STL都有, 还有prev_permutation, 实做这个 05/30 09:58
4F：→ shaopin:要做啥? 05/30 09:58
5F：→ bigpigbigpig:试试 L = [ 0, 0, 1, 1, 1 ] :) 05/30 14:58