[leetcode] Compare Version Numbers

From : https://leetcode.com/problems/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

class Solution {
public:
    int compareVersion(string version1, string version2) {
        int v1, v2, i=0, j=0, len1 = version1.size(), len2 = version2.size();
        while(i<len1 || j<len2) {
            v1 = v2 = 0;
            while(i<len1 && version1[i]!='.') v1 = v1*10 + version1[i++]-'0';
            while(j<len2 && version2[j]!='.') v2 = v2*10 + version2[j++]-'0';
            if(v1!=v2) return (v1>v2)-(v1<v2);
			i++;j++;
        }
        return 0;
    }
};

public class Solution {
    public int compareVersion(String version1, String version2) {
        for (int i = 0, j = 0; i < version1.length() || j < version2.length(); ++i, ++j) {
			int v1 = 0;
			int v2 = 0;
			while (i < version1.length() && version1.charAt(i) != '.') {
				v1 = v1 * 10 + version1.charAt(i++) - '0';
			}
			while (j < version2.length() && version2.charAt(j) != '.') {
				v2 = v2 * 10 + version2.charAt(j++) - '0';
			}
			if (v1 != v2) {
				return v1 > v2 ? 1 : -1;
			}
		}
		return 0;
    }
}