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

 

解题：

题目的意思是，比较版本号的大小，2版本比1版本大，而2.10版本比2.5版本大；

分别取出小数点前和小数点后的数字做比较；

代码：

1 class Solution { 2 public: 3     int compareVersion(string version1, string version2) { 4         int len1 = version1.size();  5         int len2 = version2.size(); 6         int num1 = 0; 7         int num2 = 0; 8         int i = 0;  9         int j = 0;10         while(i < len1 || j < len2) {11             while (i < len1 && version1[i] != '.') {12                 num1 = num1 * 10 + (version1[i] - '0');13                 i++;14             }15 16             while (j < len2 && version2[j] != '.') {17                 num2 = num2 * 10 + (version2[j] - '0');18                 j++;19             }20 21             if(num1 > num2) 22                 return 1;23             else if (num1 < num2) 24                 return -1;25 26             num1 = 0;27             num2 = 0;28             i++;29             j++;30         }31 32         return 0;33     }34 };