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:

1
0.1 < 1.1 < 1.2 < 13.37

思路

题目其实不难，需要我们去比较版本号。

要注意，版本号和小数不同，可以是诸如major.minor(.build)这样的形式。其中，build号还可以用日期（如1.2.20170101来命名，另外，版本号的排序也有讲究。举例：

1.124 > 1.4 > 0.134 > 0.5 > 0.0.2 > 0.0.1

弄清楚以上的规律，代码也就好组织了。将所提供的字符串按照规则转换为可以比较的形式即可。在代码中，我利用了list对圆点隔开的每一个数进行存储，从而也可以比较版本号。

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class Solution(object):
    def compareVersion(self, version1, version2):
        """
        :type version1: str
        :type version2: str
        :rtype: int
        """
        v1 = []
        v2 = []
        tmp = ""
        for i in version1:
            if (i == "."):
                v1.append(int(tmp))
                tmp = ""
            else:
                tmp += i
        v1.append(int(tmp))
        tmp = ""
        
        for j in version2:
            if (j == "."):
                v2.append(int(tmp))
                tmp = ""
            else:
                tmp += j
        v2.append(int(tmp))
        
        while (len(v1) and v1[-1] == 0):
            del v1[-1]
        while (len(v2) and v2[-1] == 0):
            del v2[-1]
        
        len1 = len(v1)
        len2 = len(v2)
        for i in range(min(len1,len2)):
            if (v1[i] > v2[i]):
                return 1
            elif (v1[i] < v2[i]):
                return -1
        if (len1 == len2):
            return 0
        elif (len1 > len2):
            return 1
        else:
            return -1