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

Solution

• Java: 利用compareTo和split File /Users/minli/Desktop/Personal/sdytlm.github.io/source/downloads/code/LeetCode/Java/Compare-Version-Numbers.java could not be found

• Python: 考虑清除”1.0.1” 和 “1.0”的情况

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class Solution(object):
    def compareVersion(self, version1, version2):
        """
        :type version1: str
        :type version2: str
        :rtype: int
        """


        s1 = version1.split('.')
        s2 = version2.split('.')

        i = 0
        j = 0

        while i<len(s1) and j < len(s2):
            if int(s1[i]) < int(s2[j]):
                return -1
            if int(s1[i]) > int(s2[j]):
                return 1
            i += 1
            j += 1

        while i < len(s1):
            if int(s1[i]) != 0:
                return 1
            i += 1

        while j < len(s2):
            if int(s2[j]) != 0:
                return -1
            j += 1
        return 0