题目

restore-ip-addresses/

算法

* 递归

* 包里

代码

* 递归

class Solution {
public:
    vector<string> restoreIpAddresses(string s) {
        vector<string> res;
        restore(s, 4, "", res);
        return res;
    }
    void restore(string s, int k, string out, vector<string> &res) {
        if (k == 0) {
            if (s.empty()) res.push_back(out);
        }
        else {
            for (int i = 1; i <= 3; ++i) {
                if (s.size() >= i && isValid(s.substr(0, i))) {
                    if (k == 1) restore(s.substr(i), k - 1, out + s.substr(0, i), res);
                    else restore(s.substr(i), k - 1, out + s.substr(0, i) + ".", res);
                }
            }
        }
    }
    bool isValid(string s) {
        if (s.empty() || s.size() > 3 || (s.size() > 1 && s[0] == '0')) return false;
        int res = atoi(s.c_str());
        return res <= 255 && res >= 0;
    }
};

* 暴力

class Solution {
public:
    vector<string> restoreIpAddresses(string s) {
        vector<string> res;
        for (int a = 1; a < 4; ++a) 
        for (int b = 1; b < 4; ++b) 
        for (int c = 1; c < 4; ++c) 
        for (int d = 1; d < 4; ++d) 
            if (a + b + c + d == s.size()) {
                int A = stoi(s.substr(0, 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 <= 255 && B <= 255 && C <= 255 && D <= 255) {
                    string t = to_string(A) + "." + to_string(B) + "." + to_string(C) + "." + to_string(D);
                    if (t.size() == s.size() + 3) res.push_back(t);
                }
            }
        return res;
    }
};