Check 里的“Three words”题目的解题思路是怎么样的？

[Graxce]

我的思路是 先 split ，然后判断类型。如果连续三次相等就返回 True 。

下面是题目

Let's teach the Robots to distinguish words and numbers.

You are given a string with words and numbers separated by whitespaces (one space). The words contains only letters. You should check if the string contains three words in succession. For example, the string "start 5 one two three 7 end" contains three words in succession.

Input: A string with words.

Output: The answer as a boolean.

Example:

checkio("Hello World hello") == True
checkio("He is 123 man") == False
checkio("1 2 3 4") == False
checkio("bla bla bla bla") == True
checkio("Hi") == False

How it is used: This teaches you how to work with strings and introduces some useful functions.

Precondition: The input contains words and/or numbers. There are no mixed words (letters and digits combined).
0 < len(words) < 100

http://www.checkio.org/mission/three-words/

这个是地址
大家可以说下思路，不用具体写代码。谢谢了！

[Slienc7]

regex:
([a-zA-Z]{1,99}\s[a-zA-Z]{1,99}\s[a-zA-Z]{1,99})

[Graxce]

@xgowex 唔，忘了说了。只能用内置函数。

[lijsh]

题目没说只能用内置函数啊，我也是用正则的；
高票答案里除了正则以外就是定义个 count 变量，结合`isalpha()`方法。

[Graxce]

@lijsh 可是我为什么导包会报错呢！！真是醉了。

[Graxce]

def checkio(words):
succ = 0
for word in words.split():
succ = (succ + 1)*word.isalpha()
if succ == 3: return True
else: return False


这个答案很赞，我也想到用 split 切开然后逐个判断。可是脑袋就是转不过来。