leetcode 165 Compare Version Numbers python

純屬吐槽。。那坑爹的題目，不過也有多是我e文沒看太仔細吧

原題目

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
Credits:
Special thanks to @ts for adding this problem and creating all test cases.

目測就是一個版本號比較，搞起。。。

而後就掉坑裏了。。

看測試用例，尼瑪。。 拿'0' 跟'1'比較。。。 你贏了。。

繼續。。 尼瑪 testcase 裏面有'0.0.1'

算你狠。。。 接着改

最後一個是'0.1'跟'0.0.1' 比較版本...

你大爺。。。 

最終版本。。雖然確實醜了點

最終版本

# coding = utf8


class Solution:
    # @param {string} version1
    # @param {string} version2
    # @return {integer}
    def compareVersion(self, version1, version2):
        v1 = self.compareVersionToTuple(version1)
        v2 = self.compareVersionToTuple(version2)

        deep = max(len(v1), len(v2), 2)

        while len(v1) < deep:
            v1.append(0)

        while len(v2) < deep:
            v2.append(0)

        for n in range(deep):
            sv_1 = v1[n]
            sv_2 = v2[n]
            if sv_1 > sv_2:
                return 1
            elif sv_1 < sv_2:
                return -1
        return 0


    def compareVersionToTuple(self, version):
        v = []
        substrings = version.split('.')
        for str in substrings:
            v.append(int(str))
        return v

    def __init__(self):
        ans = self.compareVersion('0.1', '0.0.1')
        print(ans)


solution = Solution()

最後pycharm大法好。。。