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 题意非常明白，就是按照version比较规则来比较。 自己没怎么想，看了别人的解法。 碰到‘.’之前一直累加数，碰到'.'进行比较，直至字符串结束。 字符串转数字s[i]-'0'。 就没有什么难点了。

1 class Solution { 2 public: 3  4     int compareVersion(string version1, string version2) { 5         int val1=0,val2=0; 6         int len1=version1.length(); 7         int len2=version2.length(); 8         int i=0,j=0; 9         10         while(i
     
      
       val2) return 1;20             if(val1