Discuss / Python / 通过迭代找出不是空格的切片操作的起始和结束位置

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

def trim(s): x = 0 y = len(s) for d in s: if d == ' ': x = x + 1 else: break for d in s[::-1]: if d == ' ': y = y - 1 else: break s = s[x:y] return s

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

为什么我直接返回s[x:y]不对啊，求解答

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

直接返回可以的呀

-- coding: utf-8 --

def trim(s): x = 0 y = len(s) for d in s: if d == ' ': x = x + 1 else: break for d in s[::-1]: if d == ' ': y = y - 1 else: break

return s[x:y]

print(trim(' hello '))

if trim(' hello ') != 'hello': print('测试失败!') else: print('测试成功！')