leetcode笔记--Compare Version Numbers

题目：难度（Easy）

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
Tags：String

分析：比较版本串，要考虑的有：1.2个串可能不等长，那自然是版本串的最前面的版本数字是”高位“，后面的是”低位“，比较时先比较”高位“能更早得出结论。不等长的版本串也可能”相等“，例如”1.2“与”1.2.0“。

代码实现：

class Solution(object):    def compareVersion(self, version1, version2):        """        :type version1: str        :type version2: str        :rtype: int        """        #将字符串拆分开并转化成int放到数组里        v1 = map(int, version1.split("."))        v2 = map(int, version2.split("."))        #为较短的list的末尾填充0，方便后面的比较        if len(v1) > len(v2):            length = len(v1)            i = len(v2)            while i < length:                v2.append(0)                i += 1        else:            length = len(v2)            i = len(v1)            while i < length:                v1.append(0)                i += 1                i = 0        while i < length:            if v1[i] > v2[i]:                return 1            elif v1[i] < v2[i]:                return -1            else:                i += 1        if i == length:            return 0