使用递归从字符列表中打印n个长度的组合

问题描述

我必须使用递归来解决此练习.

I have to solve this exercise using recursion.

对我来说，这真是令人难以置信，所有这些思考通常都是递归的.

It's very mind-blowing for me, all this thinking recursively generally.

例如，对于这个问题，我认为已经一个半小时，不知道我在想什么.我不知道如何开始递归思考，我是从哪里开始的?

For example, for this problem, I think already one and a half hour, not knowing what I'm thinking about. I don't know how to start thinking recursively, what am I starting from?

我写了些废话:

def print_sequences(char_list, n):
if len(char_list) == 1:
    print(char_list[1])
else:
    sub_str = ""
    for c in char_list:
        print_sequences(list(sub_str + c), n)

请帮助我发展递归意识.

Please help me develop some sense of recursion.

推荐答案

您非常亲密.您需要检查累积的池"列表的长度是否等于参数n，而不是len(char_list) == 1.但是，要创建一个列表以累积组合，请在函数签名中创建一个附加参数:

You are quite close. Instead of len(char_list) == 1, you need to check that the length of your accumulated "pool" list is equal to the parameter n. However, to create a list to accumulate the combinations, create one additional parameter in the signature of the function:

def print_sequences(char_list, n, _accum):
  if len(_accum) == n:
     print(_accum)
  else:
     for c in char_list:
        print_sequences(char_list, n, _accum+[c])


print_sequences([1, 2, 3, 4], 4, [])

输出:

[1, 1, 1, 1]
[1, 1, 1, 2]
[1, 1, 1, 3]
[1, 1, 1, 4]
[1, 1, 2, 1]
[1, 1, 2, 2]
[1, 1, 2, 3]
[1, 1, 2, 4]
[1, 1, 3, 1]
[1, 1, 3, 2]
[1, 1, 3, 3]
[1, 1, 3, 4]
[1, 1, 4, 1]
[1, 1, 4, 2]
[1, 1, 4, 3]
[1, 1, 4, 4]
[1, 2, 1, 1]
....

将_accum实现为默认列表:

def print_sequences(char_list, n, _accum=[]):
  if len(_accum) == n:
    print(_accum)
  else:
    for c in char_list:
      print_sequences(char_list, n, _accum+[c])

print_sequences([1, 2, 3, 4], 4)

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