这是受Haskell实现使用列表理解的启发：

def permutation(list):
    if len(list) == 0:
        return [[]]
    else:
        return [[x] + ys for x in list for ys in permutation(delete(list, x))]

def delete(list, item):
    lc = list[:]
    lc.remove(item)
    return lc

人们确实可以对每个排列的第一个元素进行迭代，正如tzwen的答案。然而，这样编写此解决方案更有效：

def all_perms(elements):
    if len(elements) <= 1:
        yield elements  # Only permutation possible = no permutation
    else:
        # Iteration over the first element in the result permutation:
        for (index, first_elmt) in enumerate(elements):
            other_elmts = elements[:index]+elements[index+1:]
            for permutation in all_perms(other_elmts): 
                yield [first_elmt] + permutation

这个解决方案大约快了30%，显然是因为递归以len（元素）<=1而不是0结尾。它的内存效率也高得多，因为它使用了一个生成器函数（通过yield），就像Riccardo Reyes的解决方案一样。

使用标准库中的itertools.permutations：

import itertools
list(itertools.permutations([1, 2, 3]))

从这里改编的是itertools.permutations如何实现的演示：

def permutations(elements):
    if len(elements) <= 1:
        yield elements
        return
    for perm in permutations(elements[1:]):
        for i in range(len(elements)):
            # nb elements[0:1] works in both string and list contexts
            yield perm[:i] + elements[0:1] + perm[i:]

itertools.permutations文档中列出了两种替代方法

def permutations(iterable, r=None):
    # permutations('ABCD', 2) --> AB AC AD BA BC BD CA CB CD DA DB DC
    # permutations(range(3)) --> 012 021 102 120 201 210
    pool = tuple(iterable)
    n = len(pool)
    r = n if r is None else r
    if r > n:
        return
    indices = range(n)
    cycles = range(n, n-r, -1)
    yield tuple(pool[i] for i in indices[:r])
    while n:
        for i in reversed(range(r)):
            cycles[i] -= 1
            if cycles[i] == 0:
                indices[i:] = indices[i+1:] + indices[i:i+1]
                cycles[i] = n - i
            else:
                j = cycles[i]
                indices[i], indices[-j] = indices[-j], indices[i]
                yield tuple(pool[i] for i in indices[:r])
                break
        else:
            return

另一个基于itertools.product：

def permutations(iterable, r=None):
    pool = tuple(iterable)
    n = len(pool)
    r = n if r is None else r
    for indices in product(range(n), repeat=r):
        if len(set(indices)) == r:
            yield tuple(pool[i] for i in indices)

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]

from typing import List
import time, random

def measure_time(func):
    def wrapper_time(*args, **kwargs):
        start_time = time.perf_counter()
        res = func(*args, **kwargs)
        end_time = time.perf_counter()
        return res, end_time - start_time

    return wrapper_time


class Solution:
    def permute(self, nums: List[int], method: int = 1) -> List[List[int]]:
        perms = []
        perm = []
        if method == 1:
            _, time_perm = self._permute_recur(nums, 0, len(nums) - 1, perms)
        elif method == 2:
            _, time_perm = self._permute_recur_agian(nums, perm, perms)
            print(perm)
        return perms, time_perm

    @measure_time
    def _permute_recur(self, nums: List[int], l: int, r: int, perms: List[List[int]]):
        # base case
        if l == r:
            perms.append(nums.copy())

        for i in range(l, r + 1):
            nums[l], nums[i] = nums[i], nums[l]
            self._permute_recur(nums, l + 1, r , perms)
            nums[l], nums[i] = nums[i], nums[l]

    @measure_time
    def _permute_recur_agian(self, nums: List[int], perm: List[int], perms_list: List[List[int]]):
        """
        The idea is similar to nestedForLoops visualized as a recursion tree.
        """
        if nums:
            for i in range(len(nums)):
                # perm.append(nums[i])  mistake, perm will be filled with all nums's elements.
                # Method1 perm_copy = copy.deepcopy(perm)
                # Method2 add in the parameter list using + (not in place)
                # caveat: list.append is in-place , which is useful for operating on global element perms_list
                # Note that:
                # perms_list pass by reference. shallow copy
                # perm + [nums[i]] pass by value instead of reference.
                self._permute_recur_agian(nums[:i] + nums[i+1:], perm + [nums[i]], perms_list)
        else:
            # Arrive at the last loop, i.e. leaf of the recursion tree.
            perms_list.append(perm)



if __name__ == "__main__":
    array = [random.randint(-10, 10) for _ in range(3)]
    sol = Solution()
    # perms, time_perm = sol.permute(array, 1)
    perms2, time_perm2 = sol.permute(array, 2)
    print(perms2)
    # print(perms, perms2)
    # print(time_perm, time_perm2)
```

另一种解决方案：

def permutation(flag, k =1 ):
    N = len(flag)
    for i in xrange(0, N):
        if flag[i] != 0:
            continue
        flag[i] = k 
        if k == N:
            print flag
        permutation(flag, k+1)
        flag[i] = 0

permutation([0, 0, 0])