Compare two version numbersversion1andversion2.
Ifversion1>version2return 1, ifversion1<version2return -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,2.5is 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
不用split("\\.") 的方法, 这题好恶心, 刚开始直接用字符串比较发现01 == 1
后来变成数字比较, 发现1.0 == 1 所以短的还需要自动补零;。。。。。