human
I found an interesting problem on https://leetcode.com/problems/compare-version-numbers/:
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.

Example 1:

Input: version1 = "0.1", version2 = "1.1"
Output: -1

Example 2:

Input: version1 = "1.0.1", version2 = "1"
Output: 1

Example 3:

Input: version1 = "7.5.2.4", version2 = "7.5.3"
Output: -1

I tried it in Python, but could not do it. Can you solve it?

 ... Give me a Python program to compare two version numbers version1 and version2?
assistant
#Here is Python program to compare two version numbers version1 and version2.

class Solution:
     def compareVersion(self, version1, version2):
         """
         :type version1: str
         :type version2: str
         :rtype: int
         """
         
         
         if version1 == version2:
             return 0
         else:
             version1_nums = version1.split('.')
             version2_nums = version2.split('.')
             len1, len2 = len(version1_nums), len(version2_nums)
             end = max(len1, len2 )
             version1_nums.extend([0]*(end - len1))
             version2_nums.extend([0]*(end - len2))
             for i in range(end):
                 if int(version1_nums[i]) < int(version2_nums[i]):
                     return -1
                 elif int(version1_nums[i]) > int(version2_nums[i]):
                     return 1
             return 0