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 < 13.37Credits:
Special thanks to @ts for adding this problem and creating all test cases.
这题主要要考虑 不等长度的情况。
我的做法是当不等长的时候在前面加0
class Solution:# @param version1, a string# @param version2, a string# @return an integerdef compareVersion(self, version1, version2):num1=version1.split('.')num2=version2.split('.')while len(num1) or len(num2):if len(num1)==0:num1=[0]elif len(num2)==0:num2=[0]else:i1=int(num1[0])i2=int(num2[0])if i1<i2:return -1elif i1>i2:return 1else:num1=num1[1:]num2=num2[1:]return 0
，赶快上路吧，不要有一天我们在对方的葬礼上说，要是当时去了就好了。
