// Source : https://oj.leetcode.com/problems/compare-version-numbers/
// Author : Hao Chen
// Date   : 2014-12-16

/********************************************************************************** 
 * 
 * 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
 * 
 * Credits:Special thanks to @ts for adding this problem and creating all test cases.
 *               
 **********************************************************************************/

class Solution {
    public:
        vector<int> &split(const string &s, char delim, vector<int> &elems) {
            stringstream ss(s);
            string item;
            while (getline(ss, item, delim)) {
                elems.push_back(atoi(item.c_str()));
            }
            return elems;
        }


        vector<int> split(const string &s, char delim) {
            vector<int> elems;
            split(s, delim, elems);
            return elems;
        }

        void rightTrimZero(vector<int> &v){
            while (v.back() == 0){
                v.pop_back();
            }
        }

        int compareVersion(string version1, string version2) {

            //split the version by delimer '.'
            vector<int> ver1 = split(version1, '.');
            vector<int> ver2 = split(version2, '.');

            //remove the right Zeros
            rightTrimZero(ver1);
            rightTrimZero(ver2);

            //compare two versions 
            for (int i=0; i<ver1.size() && i < ver2.size(); i++) {
                if (ver1[i] < ver2[i]){
                    return -1;
                }else if(ver1[i] > ver2[i]){
                    return 1;
                }
            }

            //if the above for-loop is not returned, which means they are equal so far
            //then check the length.
            if (ver1.size() > ver2.size()) {
                return 1;
            }else if (ver1.size() < ver2.size()) {
                return -1;
            }

            return 0;

        }

};