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165.py
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62 lines (44 loc) · 1.43 KB
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'''
165. 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.
Here is an example of version numbers ordering:
0.1 < 1.1 < 1.2 < 13.37
'''
class Solution(object):
def compareVersion(self, version1, version2):
"""
:type version1: str
:type version2: str
:rtype: int
"""
v1 = version1.split(".")
v2 = version2.split(".")
d = len(v2) - len(v1)
v1 += [0] * d
v2 += [0] * -d
for i in range(len(v1)):
if int(v1[i]) < int(v2[i]):
return -1
elif int(v1[i]) > int(v2[i]):
return 1
else:
continue
return 0
def fasterSolution(self, version1, version2):
v1, v2 = (map(int, v.split('.')) for v in (version1, version2))
d = len(v2) - len(v1)
return cmp(v1 + [0]*d, v2 + [0]*-d)
if __name__ == "__main__":
v = ["01",\
"1"]
sol = Solution()
print(sol.compareVersion(v[0], v[1]))