在置换问题中是否可以不使用嵌套循环？

Question

我正在编写一个打印 [a,b,c,d] 的所有排列的代码。我不想使用递归函数，而是使用了 4 个 for 循环，但我的代码的缺点是循环有与列表中的元素数量相同。我的问题是是否可以编写与元素数量无关的代码。

alphabet=["a","b","c","d"]

for first in alphabet:

    for second in alphabet:

        if second != first:

            for third in alphabet:

                if third!=second and third!=first :

                    for fourth in alphabet:

                        if fourth!=first and fourth!=second and fourth != third:         

                            print(first,second,third,fourth)

Answer 1

这是一个依赖于初始池中没有重复项的非递归尝试：from collections import dequedef perms(pool):    agenda = deque([([], pool)])    while agenda:        perm, left = agenda.popleft()        if not left:            yield perm            # Or, to mimic the original             # print(*perm)        else:            for x in left:                agenda.append((perm+[x], [y for y in left if y != x]))>>> list(perms('abc')))[['a', 'b', 'c'], ['a', 'c', 'b'], ['b', 'a', 'c'], ['b', 'c', 'a'], ['c', 'a', 'b'], ['c', 'b', 'a']]

Answer 2

这是一种基本方法（很容易想出）。先从简单开始。首先，让alpha = ["a", "b", "c", "d"]. 首先找到所有以 开头的排列"a"：    start_a = [["a"]]    two_start_a = [ start_a[0] + [i]  for i in alpha ]    to_three_start_a = [ [ j + [i] for i in alpha ] for j in two_start_a ]    three_start_a = []    for i in to_three_start_a:        #cleaned is to exclude list with multiple values        cleaned = [lst for lst in i if len(set(lst)) == len(lst)]         three_start_element += cleaned    to_four_start_a = [ [ j + [i] for i in alpha ] for j in three_start_a ]    four_start_a = []    for i in to_four_start_a:        #cleaned is to exclude list with multiple values        cleaned = [lst for lst in i if len(set(lst)) == len(lst)]         four_start_element += cleaned现在我们four_start_a包含所有以 开头的排列"a"。自动版本是    start_a = [[alpha[0]]]    two_start_a = [ start_a[0] + [i]  for i in alpha ]    k_start_element = two_start_element    for k in range(3, len(alpha)+1):         to_k_start_a = [ [ j + [i] for i in alpha ] for j in k_start_element ]        k_start_a = []        for i in to_k_start_a:            #cleaned is to exclude list with multiple values            cleaned = [lst for lst in i if len(set(lst)) == len(lst)]             k_start_element += cleaned那么最后的k_start_a就是从 的第一个元素开始的所有排列alpha。解决方案所以，对于所有的字母，我们可以自动化如下all_permutations = []for element in alpha:    start_element = [[element]]    two_start_element = [ start_element[0] + [i]  for i in alpha ]    k_start_element = two_start_element    for k in range(3, len(alpha)+1):         to_k_start_element = [ [ j + [i] for i in alpha ] for j in k_start_element ]        k_start_element = []        for i in to_k_start_element:            #to exclude list with multiple values            cleaned = [lst for lst in i if len(set(lst)) == len(lst)]             k_start_element += cleaned    all_permutations.extend( k_start_element )