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