Compare Version Numbers

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

Credits:

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

思路： 每次在‘.’之前的树转化成十进制的 比较。

public class Solution {  public int compareVersion(String version1, String version2) {       int i = 0, j = 0;              while(i < version1.length() || j < version2.length()){       int a = 0, b = 0;       while(i < version1.length() && version1.charAt(i) != '.'){   //判越界的要放在前面 否则放在后面容易报抛出异常 比如输入1,0时       a = a * 10 + (version1.charAt(i) - '0');    //注意减去'0'转换为数字       i++;       }       while(j < version2.length() && version2.charAt(j) != '.'){       b = b * 10 + (version2.charAt(j) - '0');       j++;       }        i++;       j++;             if(a > b) return 1;       if(a < b) return -1;       }       return 0;    }}