165 Compare Version Numbers

来源：Leetcode

原帖：https://leetcode.com/problems/compare-version-numbers/

题目：
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

代码：

 class Solution {  
 public:  
   int compareVersion(const string &v1, const string &v2, int i1, int i2){  
     int n1 = 0, n2 = 0;  
     if(i1 >= v1.size() && i2 >= v2.size()) return 0;  
     int j1 = i1, j2 = i2;   
     while (j1 < v1.size() && v1[j1] != '.') n1 = 10*n1 + v1[j1++]-'0';  
     while (j2 < v2.size() && v2[j2] != '.') n2 = 10*n2 + v2[j2++]-'0';  
     if (n1 > n2) return 1;  
     else if (n1 < n2) return -1;  
     else return compareVersion(v1, v2, ++j1, ++j2);  
   }  
   
   int compareVersion(string version1, string version2) {  
     return compareVersion(version1, version2, 0, 0);  
   }  
 };