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;				else					s1 = 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;				else					s2 = 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);			}		}	}}

Java-Compare Version Numbers