leetcode 165 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.

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class Solution {
public:
    int compareVersion(string version1, string version2) {
        int a1 = 0, a2 = 0, b1 = 0, b2 = 0;
		
		if(version1=="" && version2=="") return 0;
		
		int index1 = version1.find('.'), index2 = version2.find('.');

		if(index1==string::npos) {
			a1 = atoi(version1.c_str());
			index1 = version1.size()-1;
		} else {
			a1 = atoi(version1.substr(0, index1).c_str());
		}

		if(index2==string::npos) {
			a2 = atoi(version2.c_str());
			index2 = version2.size()-1;
		} else {
			a2 = atoi(version2.substr(0, index2).c_str());
		}
		
		if(a1 > a2) return 1;
		else if(a1 < a2) return -1;

		return compareVersion(version1.substr(index1+1), version2.substr(index2+1));
        
    }
};