Leetcode OJ : Compare Version Numbers Python solution

Total Accepted: 12400 Total Submissions: 83230

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

Leetcode OJ : Compare Version Numbers Python solution

原文：http://www.cnblogs.com/ydlme/p/4297145.html