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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
将两个版本号码的字符串以点分隔的方式转换为数字放入数组，如果长度不相等则在短的后面补0，最后两数组逐个元素比较。
class Solution: # @param version1, a string # @param version2, a string # @return an integer def compareVersion(self, s1, s2): s1 += ‘.‘ s2 += ‘.‘ x1, x2 = [], [] while s1 or s2: for i in range(len(s1)): if s1[i] == ‘.‘: x1.append(int(s1[:i])) break s1 = s1[i+1:] for i in range(len(s2)): if s2[i] == ‘.‘: x2.append(int(s2[:i])) break s2 = s2[i+1:] if len(x1) > len(x2): for i in range(len(x1)-len(x2)): x2.append(0) if len(x2) > len(x1): for i in range(len(x2)-len(x1)): x1.append(0) for i in range(len(x1)): if x1[i] > x2[i]: return 1 if x2[i] > x1[i]: return -1 return 0
Leetcode 165 Compare Version Numbers
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原文地址：http://www.cnblogs.com/lilixu/p/4658842.html
