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}")

生成所有可能的排列

我正在使用python3.4：

def calcperm(arr, size):
    result = set([()])
    for dummy_idx in range(size):
        temp = set()
        for dummy_lst in result:
            for dummy_outcome in arr:
                if dummy_outcome not in dummy_lst:
                    new_seq = list(dummy_lst)
                    new_seq.append(dummy_outcome)
                    temp.add(tuple(new_seq))
        result = temp
    return result

测试用例：

lst = [1, 2, 3, 4]
#lst = ["yellow", "magenta", "white", "blue"]
seq = 2
final = calcperm(lst, seq)
print(len(final))
print(final)

免责声明：无耻的插件由包作者。：）

trotter包与大多数实现的不同之处在于，它生成的伪列表实际上不包含排列，而是描述排列与排序中各个位置之间的映射，从而可以处理非常大的排列“列表”，如本演示所示，它在一个包含字母表中所有字母排列的伪列表中执行相当即时的操作和查找，而不使用比典型网页更多的内存或处理。

在任何情况下，要生成排列列表，我们可以执行以下操作。

import trotter

my_permutations = trotter.Permutations(3, [1, 2, 3])

print(my_permutations)

for p in my_permutations:
    print(p)

输出：

A pseudo-list containing 6 3-permutations of [1, 2, 3].
[1, 2, 3]
[1, 3, 2]
[3, 1, 2]
[3, 2, 1]
[2, 3, 1]
[2, 1, 3]

用递归求解，遍历元素，取第i个元素，然后问自己：“其余项目的排列是什么”，直到没有更多的元素。

我在这里解释了解决方案：https://www.youtube.com/watch?v=_7GE7psS2b4

class Solution:
    def permute(self,nums:List[int])->List[List[int]]:
        res=[]
        def dfs(nums,path):
            if len(nums)==0:
                res.append(path)
            for i in range(len(nums)):
                dfs(nums[:i]+nums[i+1:],path+[nums[i]])
        dfs(nums,[])
        return res

无论如何，我们可以使用sympy库，也支持多集合排列

import sympy
from sympy.utilities.iterables import multiset_permutations
t = [1,2,3]
p = list(multiset_permutations(t))
print(p)

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

答案的灵感来自获取numpy数组的所有排列

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