语法练习：front3

语法练习：front3

题目：front3

Given a string, we’ll say that the front is the first 3 chars of the string. If the string length is less than 3, the front is whatever is there. Return a new string which is 3 copies of the front.

front3(‘Java’) → ‘JavJavJav’
front3(‘Chocolate’) → ‘ChoChoCho’
front3(‘abc’) → ‘abcabcabc’

我的解答：

def front3(str):
  if len(str) < 3:
    return str*3
  else:
    return (str[:3]*3)
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Expected Run
front3(‘Java’) → ‘JavJavJav’ ‘JavJavJav’ OK
front3(‘Chocolate’) → ‘ChoChoCho’ ‘ChoChoCho’ OK
front3(‘abc’) → ‘abcabcabc’ ‘abcabcabc’ OK
front3(‘abcXYZ’) → ‘abcabcabc’ ‘abcabcabc’ OK
front3(‘ab’) → ‘ababab’ ‘ababab’ OK
front3(‘a’) → ‘aaa’ ‘aaa’ OK
front3(‘’) → ‘’ ‘’ OK

All Correct

标答：

def front3(str):
  # Figure the end of the front
  front_end = 3
  if len(str) < front_end:
    front_end = len(str)
  front = str[:front_end]
  return front + front + front 
  
  # Could omit the if logic, and write simply front = str[:3]
  # since the slice is silent about out-of-bounds conditions.
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