def pzip(c, seq):
    result = []
    for item in seq:
        for i in range(len(item)+1):
            result.append(item[i:]+c+item[:i])
    return result


def perm(line):
    seq = [c for c in line]
    if len(seq) <=1 :
        return seq
    else:
        return pzip(seq[0], perm(seq[1:]))

如果用户希望在列表中保留所有排列，可以使用以下代码：

def get_permutations(nums, p_list=[], temp_items=[]):
    if not nums:
        return
    elif len(nums) == 1:
        new_items = temp_items+[nums[0]]
        p_list.append(new_items)
        return
    else:
        for i in range(len(nums)):
            temp_nums = nums[:i]+nums[i+1:]
            new_temp_items = temp_items + [nums[i]]
            get_permutations(temp_nums, p_list, new_temp_items)

nums = [1,2,3]
p_list = []

get_permutations(nums, p_list)

另一种方法（无libs）

def permutation(input):
    if len(input) == 1:
        return input if isinstance(input, list) else [input]

    result = []
    for i in range(len(input)):
        first = input[i]
        rest = input[:i] + input[i + 1:]
        rest_permutation = permutation(rest)
        for p in rest_permutation:
            result.append(first + p)
    return result

输入可以是字符串或列表

print(permutation('abcd'))
print(permutation(['a', 'b', 'c', 'd']))

list2Perm = [1, 2.0, 'three']
listPerm = [[a, b, c]
            for a in list2Perm
            for b in list2Perm
            for c in list2Perm
            if ( a != b and b != c and a != c )
            ]
print listPerm

输出：

[
    [1, 2.0, 'three'], 
    [1, 'three', 2.0], 
    [2.0, 1, 'three'], 
    [2.0, 'three', 1], 
    ['three', 1, 2.0], 
    ['three', 2.0, 1]
]

from __future__ import print_function

def perm(n):
    p = []
    for i in range(0,n+1):
        p.append(i)
    while True:
        for i in range(1,n+1):
            print(p[i], end=' ')
        print("")
        i = n - 1
        found = 0
        while (not found and i>0):
            if p[i]<p[i+1]:
                found = 1
            else:
                i = i - 1
        k = n
        while p[i]>p[k]:
            k = k - 1
        aux = p[i]
        p[i] = p[k]
        p[k] = aux
        for j in range(1,(n-i)/2+1):
            aux = p[i+j]
            p[i+j] = p[n-j+1]
            p[n-j+1] = aux
        if not found:
            break

perm(5)

以下代码是给定列表的就地排列，作为生成器实现。由于它只返回对列表的引用，因此不应在生成器外部修改列表。该解决方案是非递归的，因此使用了低内存。还可以很好地处理输入列表中元素的多个副本。

def permute_in_place(a):
    a.sort()
    yield list(a)

    if len(a) <= 1:
        return

    first = 0
    last = len(a)
    while 1:
        i = last - 1

        while 1:
            i = i - 1
            if a[i] < a[i+1]:
                j = last - 1
                while not (a[i] < a[j]):
                    j = j - 1
                a[i], a[j] = a[j], a[i] # swap the values
                r = a[i+1:last]
                r.reverse()
                a[i+1:last] = r
                yield list(a)
                break
            if i == first:
                a.reverse()
                return

if __name__ == '__main__':
    for n in range(5):
        for a in permute_in_place(range(1, n+1)):
            print a
        print

    for a in permute_in_place([0, 0, 1, 1, 1]):
        print a
    print