#!/usr/bin/env python

def perm(a, k=0):
   if k == len(a):
      print a
   else:
      for i in xrange(k, len(a)):
         a[k], a[i] = a[i] ,a[k]
         perm(a, k+1)
         a[k], a[i] = a[i], a[k]

perm([1,2,3])

输出：

[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 2, 1]
[3, 1, 2]

当我交换列表的内容时，需要一个可变的序列类型作为输入。例如，烫发（list（“ball”）会起作用，而烫发（“ball”）不会起作用，因为你不能更改字符串。

这种Python实现的灵感来自Horowitz、Sahni和Rajasekeran在《计算机算法》一书中提出的算法。

为了节省您可能的搜索和实验时间，下面是Python中的非递归置换解决方案，它也适用于Numba（从0.41版开始）：

@numba.njit()
def permutations(A, k):
    r = [[i for i in range(0)]]
    for i in range(k):
        r = [[a] + b for a in A for b in r if (a in b)==False]
    return r
permutations([1,2,3],3)
[[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]

要给人留下绩效印象：

%timeit permutations(np.arange(5),5)

243 µs ± 11.1 µs per loop (mean ± std. dev. of 7 runs, 1 loop each)
time: 406 ms

%timeit list(itertools.permutations(np.arange(5),5))
15.9 µs ± 8.61 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
time: 12.9 s

因此，只有在必须从njit函数调用它时才使用此版本，否则更倾向于itertools实现。

功能性风格

def addperm(x,l):
    return [ l[0:i] + [x] + l[i:]  for i in range(len(l)+1) ]

def perm(l):
    if len(l) == 0:
        return [[]]
    return [x for y in perm(l[1:]) for x in addperm(l[0],y) ]

print perm([ i for i in range(3)])

结果：

[[0, 1, 2], [1, 0, 2], [1, 2, 0], [0, 2, 1], [2, 0, 1], [2, 1, 0]]

我使用了一种基于阶乘数系统的算法——对于长度为n的列表，您可以逐项组装每个排列，从每个阶段留下的项目中进行选择。第一项有n个选项，第二项有n-1个选项，最后一项只有一个选项，因此可以使用阶乘数系统中数字的数字作为索引。这是数字0到n-1对应于词典顺序中的所有可能的排列。

from math import factorial
def permutations(l):
    permutations=[]
    length=len(l)
    for x in xrange(factorial(length)):
        available=list(l)
        newPermutation=[]
        for radix in xrange(length, 0, -1):
            placeValue=factorial(radix-1)
            index=x/placeValue
            newPermutation.append(available.pop(index))
            x-=index*placeValue
        permutations.append(newPermutation)
    return permutations

permutations(range(3))

输出：

[[0, 1, 2], [0, 2, 1], [1, 0, 2], [1, 2, 0], [2, 0, 1], [2, 1, 0]]

此方法是非递归的，但在我的计算机上速度稍慢，xrange在n！太大，无法转换为C长整数（我的n=13）。当我需要它的时候，它已经足够了，但它远没有itertools.permutations。

def permutations(head, tail=''):
    if len(head) == 0:
        print(tail)
    else:
        for i in range(len(head)):
            permutations(head[:i] + head[i+1:], tail + head[i])

称为：

permutations('abc')

这是初始排序后生成排列的渐近最优方式O（n*n！）。

有n个！最多进行一次置换，且具有下一次置换（..），以O（n）时间复杂度运行

在3个步骤中，

找到最大的j，使a[j]可以增加以最小可行量增加a[j]找到扩展新a[0..j]的字典最少方法

'''
Lexicographic permutation generation

consider example array state of [1,5,6,4,3,2] for sorted [1,2,3,4,5,6]
after 56432(treat as number) ->nothing larger than 6432(using 6,4,3,2) beginning with 5
so 6 is next larger and 2345(least using numbers other than 6)
so [1, 6,2,3,4,5]
'''
def hasNextPermutation(array, len):
    ' Base Condition '
    if(len ==1):
        return False
    '''
    Set j = last-2 and find first j such that a[j] < a[j+1]
    If no such j(j==-1) then we have visited all permutations
    after this step a[j+1]>=..>=a[len-1] and a[j]<a[j+1]

    a[j]=5 or j=1, 6>5>4>3>2
    '''
    j = len -2
    while (j >= 0 and array[j] >= array[j + 1]):
        j= j-1
    if(j==-1):
        return False
    # print(f"After step 2 for j {j}  {array}")
    '''
    decrease l (from n-1 to j) repeatedly until a[j]<a[l]
    Then swap a[j], a[l]
    a[l] is the smallest element > a[j] that can follow a[l]...a[j-1] in permutation
    before swap we have a[j+1]>=..>=a[l-1]>=a[l]>a[j]>=a[l+1]>=..>=a[len-1]
    after swap -> a[j+1]>=..>=a[l-1]>=a[j]>a[l]>=a[l+1]>=..>=a[len-1]

    a[l]=6 or l=2, j=1 just before swap [1, 5, 6, 4, 3, 2] 
    after swap [1, 6, 5, 4, 3, 2] a[l]=5, a[j]=6
    '''
    l = len -1
    while(array[j] >= array[l]):
        l = l-1
    # print(f"After step 3 for l={l}, j={j} before swap {array}")
    array[j], array[l] = array[l], array[j]
    # print(f"After step 3 for l={l} j={j} after swap {array}")
    '''
    Reverse a[j+1...len-1](both inclusive)

    after reversing [1, 6, 2, 3, 4, 5]
    '''
    array[j+1:len] = reversed(array[j+1:len])
    # print(f"After step 4 reversing {array}")
    return True

array = [1,2,4,4,5]
array.sort()
len = len(array)
count =1
print(array)
'''
The algorithm visits every permutation in lexicographic order
generating one by one
'''
while(hasNextPermutation(array, len)):
    print(array)
    count = count +1
# The number of permutations will be n! if no duplicates are present, else less than that
# [1,4,3,3,2] -> 5!/2!=60
print(f"Number of permutations: {count}")