[LintCode] Restore IP Address 复原IP地址

Given a string containing only digits, restore it by returning all possible valid IP address combinations.

Have you met this question in a real interview?

Yes

Example

Given "25525511135", return

[
  "255.255.11.135",
  "255.255.111.35"
]

Order does not matter.

LeetCode上的原题，请参见我之前的博客Restore IP Addresses。

解法一：

class Solution {
public:
    /**
     * @param s the IP string
     * @return All possible valid IP addresses
     */
    vector<string> restoreIpAddresses(string& s) {
        vector<string> res;
        helper(s, , "", res);
        return res;
    }
    void helper(string s, int k, string out, vector<string>& res) {
        if (k == ) {
            if (s.empty()) res.push_back(out);
            return;
        }
        for (int i = ; i < ; ++i) {
            if (s.size() < i) break;
            int val = stoi(s.substr(, i));
            if (val >  || i != to_string(val).size()) continue;
            helper(s.substr(i), k - , out + s.substr(, i) + (k ==  ? "" : "."), res);
        }
    }
};

解法二：

class Solution {
public:
    /**
     * @param s the IP string
     * @return All possible valid IP addresses
     */
    vector<string> restoreIpAddresses(string& s) {
        vector<string> res;
        for (int a = ; a < ; ++a)
        for (int b = ; b < ; ++b)
        for (int c = ; c < ; ++c)
        for (int d = ; d < ; ++d)
            if (a + b + c + d == s.size()) {
                int A = stoi(s.substr(, a));
                int B = stoi(s.substr(a, b));
                int C = stoi(s.substr(a + b, c));
                int D = stoi(s.substr(a + b + c, d));
                if (A <=  && B <=  && C <=  && D <= ) {
                    string t = to_string(A) + "." + to_string(B) + "." + to_string(C) + "." + to_string(D);
                    if (t.size() == s.size() + ) res.push_back(t);
                }
            }
        return res;
    }
};