Discuss / Python / 有没有精简的办法。。。(ps:好多同学是是判断s[0],那如果有两个空格在前面或者后面测试就失败了呀)

有没有精简的办法。。。(ps:好多同学是是判断s[0],那如果有两个空格在前面或者后面测试就失败了呀)

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def trim(s):            if s=='':                return s            for k in range (len(s)):                if s==' '*len(s):                    s=''                    return s            if s[0]!=' ':                if s[-1]!=' ':                    pass                if s[-1]==' ':                    for i in range(1,len(s)):                        if s[-i]!=' ':                            return s[:-i+1]            if s[0]==' ':                if s[-1]!=' ':                    for i in range(1, len(s)):                        if s[i]!=' ':                            return s[i:]                if s[-1]==' ':                    for i in range(len(s)):                        if s[i]!=' ':                            s_1=s[i:]                            for j in range (1,len(s_1)):                                if s_1[-j]!=' ':                                    return s_1[:-j+1]            else:                s=''                return s

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