Problem

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

Note

看到一个巧妙的解法，分享一下：
将version1，version2都转换为char array，然后分别用两个指针s1, i1和s2, i2展开循环。
首先找整数部分的坐标段，s1和s2都指向初值0，令i1和i2一直向后遍历到小数点为止。
然后用Integer.valueOf(someStr.substring(s1, i1))将version1, version2的整数段转化为数值，进行比较：若结果为大于或小于关系，直接返回结果；若结果为相等，进行小数部分的比较。
比较小数大小，只比高位即可，所以可以沿用这个循环，只需要继续用i1和i2向后移动，同时赋坐标值给s1和s2。这样转化数值函数中的小数段每次遍历就只有一个字符：str.substring(i1, i1)，即str.charAt(i1)。

Solution

public class Solution {
    public int compareVersion(String version1, String version2) {
    char[] v1 = version1.toCharArray();
    char[] v2 = version2.toCharArray();
    int i1 = 0,i2 = 0;
    int s1 = 0,s2 = 0;
    while(i1 < v1.length || i2 < v2.length){
        while(i1 < v1.length && v1[i1] != '.') i1++;
        while(i2 < v2.length && v2[i2] != '.') i2++;
        int a = s1 < v1.length ? Integer.valueOf(version1.substring(s1,i1)) : 0;
        int b = s2 < v2.length ? Integer.valueOf(version2.substring(s2,i2)) : 0;
        if(a > b) return 1;
        if(a < b) return -1;
        s1 = ++i1;
        s2 = ++i2;
    }
    return 0;
    }
}