LeetCode OJ 系列之165 Compare Version Numbers --Python

Problem:

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

Answer:

class Solution(object):    def compareVersion(self, version1, version2):        """        :type version1: str        :type version2: str        :rtype: int        """        tmp1 = version1.split('.')        tmp2 = version2.split('.')                        if len(tmp1)<len(tmp2):            for i in range(len(tmp1),len(tmp2)):                tmp1.append('0')        elif len(tmp1)>len(tmp2):            for i in range(len(tmp2),len(tmp1)):                tmp2.append('0')        for i in range(len(tmp1)):            if int(tmp1[i])>int(tmp2[i]): return 1            if int(tmp1[i])<int(tmp2[i]): return -1        return 0