Java-Compare Version Numbers

Compare two version numbers version1 and version1.
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

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

比较版本号的新旧

以 . 为界  依次比较每个 . 前的数大小 分出大小就输出 相等就继续 特殊情况是类似 1与1.0的 此时也认为他们相等 代码如下;

public class Solution {    public static int compareVersion(String version1, String version2) {return compare(version1, 0, version2, 0);}    public static int compare(String version1,int st1, String version2,int st2) {int ssa = 1;int ssb = 1;if (st1 == version1.length() || st1 == version1.length() + 1) {ssa = 0;if (st2 == version2.length() || st2 == version2.length() + 1) {return 0;} }if (st2 == version2.length() || st2 == version2.length() + 1) {ssb = 0;if (st1 == version1.length() || st1 == version1.length() + 1) {return 0;} }int i = st1;int a=0;if (ssa != 0) {String s1 = "";for (; i < version1.length(); i++) {if (version1.charAt(i) == '.')break;elses1 = s1 + version1.charAt(i);}a = Integer.valueOf(s1);}int j = st2;int b=0;if (ssb != 0) {String s2 = "";for (; j < version2.length(); j++) {if (version2.charAt(j) == '.')break;elses2 = s2 + version2.charAt(j);}b = Integer.valueOf(s2);}if (a > b)return 1;else {if (a < b)return -1;else {if(ssa==0||ssb==0)return 0;else return compare(version1, i + 1, version2, j + 1);}}}}