这题乍一看有点像Decode Ways ，实际上是一个深搜+剪枝的题目

也可以通过三个for循环寻找可行的‘.’的位置

递归方法如下：

    vector<string> restoreIpAddresses(string s) {
        vector<string> res;
        restore(s, 0, 0, res, "");
        return res;
    }
    
    void restore(string & s, int start, int ipNum, vector<string> & res, string ip)
    {
        if (start == s.size() && ipNum == 4)
        {
            res.push_back(ip.substr(1));
            return;
        }
        if (start >= s.size() || ipNum >= 4) return;
        if (s[start] == '0')
            return restore(s, start + 1, ipNum + 1, res, ip + "." + s.substr(start, 1));
        if (s[start] == '1' || s[start] == '2')
        {
            if (!(s[start] == '2' && start + 1 < s.size() && (s[start + 1] > '5' || s[start + 1] == '5' && start + 2 < s.size() && s[start + 2] > '5')))
            restore(s, start + 3, ipNum + 1, res, ip + "." + s.substr(start, 3));
        }
        restore(s, start + 1, ipNum + 1, res, ip + "." + s.substr(start, 1));
        restore(s, start + 2, ipNum + 1, res, ip + "." + s.substr(start, 2));
    }

Restore IP Addresses [leetcode]