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

思路：题目不难，细节需要注意。具体做法，以”.”将字符串切分为几个字符串片段（split，c++木有这个函数，自己实现了一个），然后将每一个片段转换成对应的整数，最后从左至右比较对应整数的大小从而返回不同的值。需要注意的是1.0与1与1.0.0.0是相等的，代码如下：

class Solution {public:    bool isAllZero(string s){        for(int i = 0; i < s.length();++i)            if(s[i] != '0')                return false;        return true;    }    vector<string> split(string str, string s) {        vector<string> ret;        int f = str.find(s);        if (f == -1) {            ret.push_back(str);            return ret;        }        else {            int a = str.length() - f - 1;            string t = str.substr(f + 1, str.length() - f - 1);            ret = split(t, s);            ret.push_back(str.substr(0, f));        }        return ret;    }    int compareVersion(string version1, string version2) {        vector<string>v1 = split(version1, ".");        vector<string>v2 = split(version2, ".");        int i = v1.size() - 1;        int j = v2.size() - 1;        while (i >= 0 && j >= 0) {            if (atoi(v1[i].c_str()) > atoi(v2[j].c_str()))                return 1;            else if (atoi(v1[i].c_str()) < atoi(v2[j].c_str()))                return -1;            else {                i--;                j--;            }        }        while(i >= 0){            if(!isAllZero(v1[i--]))                return 1;        }        while(j >= 0){            if(!isAllZero(v2[j--]))                return -1;        }        return 0;    }};