没有 itertools 的两个值的排列使用递归！

问题描述

Stackoverflow，我再次请求您的帮助。

我知道还有其他主题与此相关，但我会解释是什么让我的作业与众不同。

基本上我的函数会得到一个 0 和 1 的列表，并返回字符串的所有可能顺序。例如，对于“0111”，我们将得到“0111”、“1011”、“1101”、“1110”。

这是我的代码：

def permutations(string):
    if len(string) == 1:
        return [string]

    lst = []

    for j in range(len(string)):
        remaining_elements = ''.join([string[i] for i in range(len(string)) if i != j])
        mini_perm = permutations(remaining_elements)

        for perm in mini_perm:
            new_str = string[j] + perm
            if new_str not in lst:
                lst.append(new_str)

    return lst

问题是当我运行像“000000000011”这样的字符串时，它需要很长时间来处理。应该有一种更有效的方法来做到这一点，因为它只是两个数字。所以我不应该使用索引？

如果你能找到一个更有效的说法，请帮助我。

（我可以使用循环，但也必须使用递归！）

解决方法

以下是使用更高效的递归创建排列的示例：

def permute(string):
    string = list(string)
    n = len(string)

    # Base conditions
    # If length is 0 or 1,there is only 1 permutation
    if n in [0,1]:
        return [string]

    # If length is 2,then there are only two permutations
    # Example: [1,2] and [2,1]
    if n == 2:
        return [string,string[::-1]]

    res = []
    # For every number in array,choose 1 number and permute the remaining
    # by calling permute recursively
    for i in range(n):
        permutations = permute(string[:i] + string[i+1:])
        for p in permutations:
            res.append([''.join(str(n) for n in [string[i]] + p)])

    return res

这也适用于 permute('000000000011') - 希望它有所帮助！

,

您还可以将 collections.Counter 与递归生成器函数一起使用：

from collections import Counter
def permute(d):
   counts = Counter(d)
   def get_permuations(c,s = []):
     if len(s) == sum(counts.values()):
        yield ''.join(s)
     else:
        for a,b in c.items():
           for i in range(1,b+1):
              yield from get_permuations({**c,a:b - i},s+([a]*i))
   return list(set(get_permuations(counts)))

print(permute("0111"))
print(permute("000000000011"))

输出：

['0111','1110','1101','1011']
['010000100000','100000000001','010000001000','000000100001','011000000000','100000000010','001001000000','000000011000','100000001000','100000100000','100001000000','001000100000','100010000000','000000001100','000100000100','010010000000','000000000011','000000100010','101000000000','110000000000','100000010000','000100001000','000001001000','000000000101','000000100100','010000000001','001000000100','001000000010','000110000000','000011000000','000001100000','000000110000','001000000001','000010001000','000100100000','000001000001','000010000001','001100000000','000100000001','001000001000','010000000100','010000010000','000000010001','001000010000','010001000000','100000000100','100100000000','000000001001','010100000000','000010100000','010000000010','000000001010','000010000100','001010000000','000000010010','000001000010','000100000010','000101000000','000000010100','000100010000','000000000110','000001000100','000010010000','000000101000','000001010000','000010000010']

,

发布某人给我的答案。感谢您的回复！：

def permutations(zeroes,ones,lst,perm):
    if zeroes == 0 and ones == 0:
        lst.append(perm)
        return
    elif zeroes < 0 or ones < 0:
        return
    
    permutations(zeroes - 1,perm + '0')
    permutations(zeroes,ones - 1,perm + '1')