[leetcode_165] Compare Version Numbers

题意的换一种说法:
将输入的两个数字,分别用 . 分割,然后逐一比较大小。如果数字数量不一样,多的大。前提是多出来的有不为 0 的。

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class Solution {
public:
    void getNums(string version1, vector<int> &v1)
    {
        int s = 0;
        int pos = version1.find(".");
        while (-1 != pos)
        {
            v1.push_back(atoi(version1.substr(s, pos - s).c_str()));
            s = pos + 1;
            pos = version1.find(".", pos + 1);
        }
        v1.push_back(atoi(version1.substr(s).c_str()));
    }
    int compareVersion(string version1, string version2) {
        vector<int> v1, v2;
        v1.clear();
        v2.clear();
        getNums(version1, v1);
        getNums(version2, v2);
        int min = v1.size() > v2.size() ? v2.size() : v1.size();
        for (int i = 0; i < min;i++)
        {
            if (v1[i] > v2[i])
            {
                return 1;
            }
            if (v1[i] < v2[i])
            {
                return -1;
            }
        }
        if (v1.size() > v2.size())
        {
            for (int i = min; i < v1.size(); i++)
            {
                if (v1[i] > 0)
                {
                    return 1;
                }
            }
            return 0;
        }
        if (v1.size() < v2.size())
        {
            for (int i = min; i < v2.size(); i++)
            {
                if (v2[i] > 0)
                {
                    return -1;
                }
            }
            return 0;
        }
        return 0;
    }
};
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