【Leetcode】Compare version numbers

【题目】

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

题目的话呢，就是让比较版本号 ， 什么 1.0.3 >1.0.2

【思路】

真的好聪明哟~

每个.之间的区域是需要比较的区域。

怎么将这些String分开：split : "//."

怎么解决长度不一样的问题：补零。

注意：！！！在java里面如果要slipt "."需要加上“//."！！！！！！

String - > integer: Integer.parseInt(String num);

I checked other Java solution and the basic idea is the same. In addition, I simply the logic by making the two version number same length. For example, if version1 = "1.0.2", and version2 = "1.0", the I will convert the version2 to "1.0.0".

【代码】

public int compareVersion(String version1, String version2) {        String[] v1 = version1.split("\\.");    String[] v2 = version2.split("\\.");        for ( int i = 0; i < Math.max(v1.length, v2.length); i++ ) {        int num1 = i < v1.length ? Integer.parseInt( v1[i] ) : 0;        int num2 = i < v2.length ? Integer.parseInt( v2[i] ) : 0;        if ( num1 < num2 ) {            return -1;        } else if ( num1 > num2 ) {            return +1;        }    }         return 0;}