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.
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
Example 4:
Input: version1 = "1.01", version2 = "1.001"
Output: 0
Explanation: Ignoring leading zeroes, both “01” and “001" represent the same number “1”
Example 5:
Input: version1 = "1.0", version2 = "1.0.0"
Output: 0
Explanation: The first version number does not have a third level revision number, which means its third level revision number is default to "0"
solution
这个题的思路就是把2个字符串按.分割逐步比较，同时处理0的情况。
class Solution:
def compareVersion(self, version1: str, version2: str) -> int:
if len(version1)==0 and len(version2)==0:
return 0
s1 = version1.split('.')[0]
s2 =version2.split('.')[0]
i1= i2 = None
if s1 !='':
i1 = int(s1)
else:
i1=0
if s2 != '':
i2 = int(s2)
else:
i2=0
if i1>i2:
return 1
elif i1<i2:
return -1
else:
return self.compareVersion(version1[len(s1)+1:],version2[len(s2)+1:])
