题目描述：

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

解题思路：

很多细节，代码比较冗余

class Solution:

    # @param version1, a string

    # @param version2, a string

    # @return an integer

    def compareVersion(self, version1, version2):

        v1 = version1.split('.')

        v2 = version2.split('.')

        l1 = len(v1)

        l2 = len(v2)

        if l1 < l2:

            for i in range(l1):

                if int(v1[i]) < int(v2[i]):

                    return -1

                elif int(v1[i]) > int(v2[i]):

                    return 1

            for i in range(l1,l2):

                if int(v2[i]) > 0:

                    return -1

            return 0

        else:

            for i in range(l2):

                if int(v1[i]) < int(v2[i]):

                    return -1

                elif int(v1[i]) > int(v2[i]):

                    return 1

            for i in range(l2,l1):

                if int(v1[i]) > 0:

                    return 1

            return 0