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递归去空格练习

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與生懼來

#1 Created at ... [Delete] [Delete and Lock User]

    if s[:1] == ' ':

        return trim(s[1:])

    elif s[-1:] == ' ':

        return trim(s[:-1])

    else:

        return s

非默幕

#2 Created at ... [Delete] [Delete and Lock User]

很牛啊,递归操作!

大佬,借鉴你的答案了。

TN

#4 Created at ... [Delete] [Delete and Lock User]

有一个问题请教一下·,当s = ' '时,会不会进入if-elif的循环呢?如果进入循环,s[1:]和s[:-1]的序号都超出了s的长度,输出的是什么?(测试了一下,可以输出不会报错)


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