题目
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.
You may assume the default revision number for each level of a version number to be 0. For example, version number 3.4 has a revision number of 3 and 4 for its first and second level revision number. Its third and fourth level revision number are both 0.
解题思路
想通过 . split，将两个版本的每一段，转成int进行比较，如果一个版本号长与另一个，则短的部分均为0
代码
class Solution:
def compareVersion(self, version1: str, version2: str) -> int:
v1, v2 = version1.split('.'), version2.split('.')
l1, l2 = len(v1), len(v2)
for i in range(max(l1, l2)):
p1 = int(v1[i]) if i < l1 else 0
p2 = int(v2[i]) if i < l2 else 0
if p1 > p2:
return 1
elif p1 < p2:
return -1
return 0
