Leetcode OJ : Compare Version Numbers Python solution

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　　Compare two version numbers version1 and version2.
If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0.
　　You may assume that the version strings are non-empty and contain only digits and the . character.
The . character does not represent a decimal point and is used to separate number sequences.
For instance, 2.5 is not "two and a half" or "half way to
version three", it is the fifth second-level revision of the second
first-level revision.
　　

Here is an example of version numbers ordering:

0.1 < 1.1 < 1.2 < 13.37
　　Credits:
Special thanks to @ts for adding this problem and creating all test cases.
　　
　　

Solution:



1 class Solution:
2     # @param version1, a string
3     # @param version2, a string
4     # @return an integer
5     def compareVersion(self, version1, version2):
6         splited1, splited2 = version1.split('.'), version2.split('.')
7         diff = len(splited1) - len(splited2)
8         
9         ext = splited1 if diff < 0 else splited2;
10         ext.extend(['0' for i in range(abs(diff))])
11         
12         for a, b in zip(splited1, splited2):
13             ret = cmp(int(a), int(b))
14             if ret != 0:
15                 return ret
16         return 0