Leetcode OJ : Compare Version Numbers Python solution
Total Accepted: 12400 Total Submissions: 83230Compare 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
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
Solution:
1 class Solution:
2 # @param version1, a string
3 # @param version2, a string
4 # @return an integer
5 def compareVersion(self, version1, version2):
6 splited1, splited2 = version1.split('.'), version2.split('.')
7 diff = len(splited1) - len(splited2)
8
9 ext = splited1 if diff < 0 else splited2;
10 ext.extend(['0' for i in range(abs(diff))])
11
12 for a, b in zip(splited1, splited2):
13 ret = cmp(int(a), int(b))
14 if ret != 0:
15 return ret
16 return 0
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