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:]))

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:]))

def permutate(l):
    for i, x in enumerate(l):
        for y in l[i + 1:]:
            yield x, y


if __name__ == '__main__':
    print(list(permutate(list('abcd'))))
    print(list(permutate([1, 2, 3, 4])))

#[('a', 'b'), ('a', 'c'), ('a', 'd'), ('b', 'c'), ('b', 'd'), ('c', 'd')]
#[(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)]

常规实现（无收益-将在内存中完成所有操作）：

def getPermutations(array):
    if len(array) == 1:
        return [array]
    permutations = []
    for i in range(len(array)): 
        # get all perm's of subarray w/o current item
        perms = getPermutations(array[:i] + array[i+1:])  
        for p in perms:
            permutations.append([array[i], *p])
    return permutations

产量实施：

def getPermutations(array):
    if len(array) == 1:
        yield array
    else:
        for i in range(len(array)):
            perms = getPermutations(array[:i] + array[i+1:])
            for p in perms:
                yield [array[i], *p]

基本思想是在第一个位置遍历数组中的所有元素，然后在第二个位置遍历所有其他元素，而没有为第一个位置选择的元素，等等。您可以使用递归来实现这一点，其中停止条件是到达一个由1个元素组成的数组，在这种情况下，您返回该数组。

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)

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}")