[LeetCode]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

version number,顾名思义就是版本号的意思啦。它题目中给的例子不太直观，我自己举一个例子就是:

1.2 < 1,10

因为1.10和1.1是不一样的，1.10代表the tenth revision,而1.1代表the first revision。所以这道题要将version string按照“.”分段，然后逐段比较。
有一点需要注意的是，一开始我的代码wrong answer，看到它卡在测试案例

“1.1”,”1.01.0”

对两个version string预处理后，得到的两组数据分别为：1,1和1,1,0。其实这种情况是相等的，但是一开始的代码中，数组没有初始化，array1[2]为负数，小于0，所以返回-1。修改最后的AC代码为：

class Solution {
public:
    vector<string> split(string s1,char c){
        vector<string> ans;
        string::size_type pos1, pos2;
        pos2= s1.find(c);
        pos1 = 0;
        while (string::npos != pos2){
            ans.push_back(s1.substr(pos1,pos2-pos1));
            pos1 = pos2 + 1;
            pos2 = s1.find(c, pos1);
        }
        if (pos1 != s1.length()){
            ans.push_back(s1.substr(pos1));
        }
        return ans;
    }
    int compareVersion(string version1, string version2) {
        vector<string> s1 = split(version1, ‘.‘);
        vector<string> s2 = split(version2, ‘.‘);
        int maxLen = s1.size()>s2.size() ? s1.size() : s2.size();
        int* array1 = new int[maxLen];
        int* array2 = new int[maxLen];
        memset(array1, 0, maxLen*sizeof(int));
        memset(array2, 0, maxLen*sizeof(int));
        int i = 0;
        for (i = 0; i < s1.size(); i++){
            array1[i] = atoi(s1[i].c_str());
        }
        for (i = 0; i < s2.size(); i++){
            array2[i] = atoi(s2[i].c_str());
        }
        for (i = 0; i < maxLen; i++){
            if (array1[i]>array2[i])
                return 1;
            else if (array1[i] < array2[i])
                return -1;
        }
        return 0;
    }
};