Question

这与递归有关。 s是字符串“ abc”。

返回s的所有排列。因此，所需的输出为：['abc'，'acb'，'bac'，'bca'，'cab'，'cba']。但是我很难理解下面的代码行：

"for perm in permute(s[:i] + s[i+1:]):"

在"s[:i] + s[i+1:]"的第一个打印语句中，它打印出的是“ c”而不是“ bc”。我以为，由于索引i从0开始（其中i = 0且让=“ a”），因此“ s [：0] + s [o + 1：]”将变为s [1：]

应返回“ bc”，因为“ b”在索引1处，而c是最后一个字母。但是打印语句返回了“ c”。

def permute(s):
    out = []

    # Base Case
    if len(s) == 1:
        out = [s]

    else:
        # For every letter in string
        for i, let in enumerate(s):

            # For every permutation resulting from Step 2 and 3 described above
            for perm in permute(s[:i] + s[i+1:]):
                print('s is ' + s)
                print('s[:i] + s[i+1:] is ' + str(s[:i] + s[i+1:]))
                print('i is ' + str(i))
                print('Current letter is ' + let)
                print('Current perm is ' + perm)

                # Add it to output
                out += [let + perm]
                print('out is ' + str(out))
                print()

    return out          

permute('abc')

s is bc
s[:i] + s[i+1:] is c
i is 0
Current letter is b
Current perm is c
out is ['bc']

s is bc
s[:i] + s[i+1:] is b
i is 1
Current letter is c
Current perm is b
out is ['bc', 'cb']

s is abc
s[:i] + s[i+1:] is bc
i is 0
Current letter is a
Current perm is bc
out is ['abc']

s is abc
s[:i] + s[i+1:] is bc
i is 0
Current letter is a
Current perm is cb
out is ['abc', 'acb']

s is ac
s[:i] + s[i+1:] is c
i is 0
Current letter is a
Current perm is c
out is ['ac']

s is ac
s[:i] + s[i+1:] is a
i is 1
Current letter is c
Current perm is a
out is ['ac', 'ca']

s is abc
s[:i] + s[i+1:] is ac
i is 1
Current letter is b
Current perm is ac
out is ['abc', 'acb', 'bac']

s is abc
s[:i] + s[i+1:] is ac
i is 1
Current letter is b
Current perm is ca
out is ['abc', 'acb', 'bac', 'bca']

s is ab
s[:i] + s[i+1:] is b
i is 0
Current letter is a
Current perm is b
out is ['ab']

s is ab
s[:i] + s[i+1:] is a
i is 1
Current letter is b
Current perm is a
out is ['ab', 'ba']

s is abc
s[:i] + s[i+1:] is ab
i is 2
Current letter is c
Current perm is ab
out is ['abc', 'acb', 'bac', 'bca', 'cab']

s is abc
s[:i] + s[i+1:] is ab
i is 2
Current letter is c
Current perm is ba
out is ['abc', 'acb', 'bac', 'bca', 'cab', 'cba']

Answer 1

这是一个烹饪实践问题。所以...

for i, let in enumerate(s): 给第一个值0 a，然后将其传递给Permute进行循环...

[0 a，1 b，2 c]这些是此时循环中的值

"for perm in permute(s[:i] + s[i+1:]):

它在您调用函数permute（）时正确，它通过了bc 它将第一个值再次调用enumerate（'bc'）0 b，然后将其传递给... permute

[0 b，1 c]这些是此时循环中的值

"for perm in permute(s[:i] + s[i+1:]):

看看发生了什么事？它在... c上调用permute（）函数！这是我们试图达到的目标。 c是第一个烫发，因为从一开始就将len（s）== 1加上进去。绝对不是一个简单的问题。编辑：

permute函数获取字符串s并将其分割为s的开始部分，直到i索引即s [：i]，其他部分为s [i + 1：]即i + 1到字符串的末尾。因此，在i为0的bc上，它需要s [：0]空字符串和s [0 + 1：]，即索引1到最后一个只是c且在其上排列的结尾。 s现在是长度1，这意味着它存储在外面。我是零abc-> bc我是零bc-> c。我是0，让= b和perm = c，那么当我= 1时，让c和perm = b。

我想到的可能会使您更好地理解此代码的东西。您应该在这样的位置添加打印语句。

    # Base Case
if len(s) == 1:
    print('Length is 1 s=',s)
    out = [s]

我认为，因为省略了它，所以它变得很难看到何时到达需要调用置换的末尾。

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