Question

所以我正在使用递归，

问题是：

" Good(43) Morning"

我必须使用递归来打印括号内的所有信息，如下所示："(43)"

到目前为止我的代码是：

def extractor(myString):
    if len(myString) == 0:
        return ""



    if myString[0] == "(":
        return myString[0]
    if myString[0] == ")":
        return myString[:]
    else:
        return extractor(myString[1:])

我只能得到括号。我怎么能改变它？

Answer 1

你快到了：

def extractor(myString):
    if not myString:  # empty strings are falsy
        return ""
    if myString[0] == "(":
        if myString[-1] == ")": # success
            return myString
        else:  # only starts with (, trim from the end
            return extractor(myString[:-1])
    else:  # doesn't start with (, trim from beginning
        return extractor(myString[1:])

Answer 2

如果您感兴趣，这是代码的迭代版本。：）

def iterative_extractor(mystring):
    if not mystring:
        return ""

    while mystring[0] != "(" or mystring[-1] != ")":
        if mystring[0] != "(":
            mystring = mystring[1:]
        else:
            mystring = mystring[:-1]
    return mystring

OR：

def iterative_extractor(mystring):
    if not mystring:
        return ""

    while mystring[0] != "(":
        mystring = mystring[1:]

    while mystring[-1] != ")":
        mystring = mystring[:-1]

    return mystring

输出：

mystring = "Lol (234) mate!"
print(iterative_extractor(mystring))
>>> "(234)"

如何使用递归选择字符串中的特定部分？

2 个答案: