V2EX = way to explore
V2EX 是一个关于分享和探索的地方
Sign Up Now
For Existing Member Sign In
推荐学习书目
› Learn Python the Hard Way
Python Sites
› PyPI - Python Package Index
› http://diveintopython.org/toc/index.html
› Pocoo
值得关注的项目
› PyPy
› Celery
› Jinja2
› Read the Docs
› gevent
› pyenv
› virtualenv
› Stackless Python
› Beautiful Soup
› 结巴中文分词
› Green Unicorn
› Sentry
› Shovel
› Pyflakes
› pytest
Python 编程
› pep8 Checker
Styles
› PEP 8
› Google Python Style Guide
› Code Style from The Hitchhiker's Guide
This topic created in 3715 days ago, the information mentioned may be changed or developed.
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！
 |
1
Slienc7 Feb 27, 2016  1
regex:
([a-zA-Z]{1,99}\s[a-zA-Z]{1,99}\s[a-zA-Z]{1,99}) |
 |
3
lijsh Feb 28, 2016 1
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。 |
 |
5
Graxce OP 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。 |
Select Language