[範例] next_permutation

[bigpigbigpig (To littlepig with love)]

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)

執行範例：

>>> 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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※ 文章網址: http://www.ptt.cc/bbs/Python/M.1401332253.A.A92.html
※ 編輯: bigpigbigpig (1.169.131.239), 05/29/2014 12:47:30
推 tiefblau:itertool表示
→ tjjh89017:這是要實作離散數學中的作法吧XD
推 shaopin:so..? C++STL都有, 還有prev_permutation, 實做這個
→ shaopin:要做啥?
→ bigpigbigpig:試試 L = [ 0, 0, 1, 1, 1 ] :)