这里有一个算法，它在不创建新的中间列表的情况下处理列表，类似于Ber在https://stackoverflow.com/a/108651/184528.

def permute(xs, low=0):
    if low + 1 >= len(xs):
        yield xs
    else:
        for p in permute(xs, low + 1):
            yield p        
        for i in range(low + 1, len(xs)):        
            xs[low], xs[i] = xs[i], xs[low]
            for p in permute(xs, low + 1):
                yield p        
            xs[low], xs[i] = xs[i], xs[low]

for p in permute([1, 2, 3, 4]):
    print p

您可以在这里亲自尝试代码：http://repl.it/J9v

此解决方案实现了一个生成器，以避免在内存中保留所有排列：

def permutations (orig_list):
    if not isinstance(orig_list, list):
        orig_list = list(orig_list)

    yield orig_list

    if len(orig_list) == 1:
        return

    for n in sorted(orig_list):
        new_list = orig_list[:]
        pos = new_list.index(n)
        del(new_list[pos])
        new_list.insert(0, n)
        for resto in permutations(new_list[1:]):
            if new_list[:1] + resto <> orig_list:
                yield new_list[:1] + resto

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。

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)
```

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

对于性能，一个由Knuth启发的numpy解决方案（第22页）：

from numpy import empty, uint8
from math import factorial

def perms(n):
    f = 1
    p = empty((2*n-1, factorial(n)), uint8)
    for i in range(n):
        p[i, :f] = i
        p[i+1:2*i+1, :f] = p[:i, :f]  # constitution de blocs
        for j in range(i):
            p[:i+1, f*(j+1):f*(j+2)] = p[j+1:j+i+2, :f]  # copie de blocs
        f = f*(i+1)
    return p[:n, :]

复制大量内存可节省时间-它比列表（itertools.permutations（range（n））快20倍：

In [1]: %timeit -n10 list(permutations(range(10)))
10 loops, best of 3: 815 ms per loop

In [2]: %timeit -n100 perms(10) 
100 loops, best of 3: 40 ms per loop