1. 题目

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

For example:
Given “25525511135”,

return [“255.255.11.135”, “255.255.111.35”]. (Order does not matter)

2. 思路

一共要提取出4个位域来。递归提取，每次从剩下的里面提取需要的个数，再讲当前合法的拼接上。
初始化点时当只需要提取一个时，直接用剩下的全部字符串匹配是否是有效的。

注：前导0的形式是在题目里是认为不合法的，比如某一位是0、1、12都是ok的，但是00、01、001、012之类的是非法的。

3. 代码

class Solution {
public:
    vector<string> restoreIpAddresses(string s) {
        return getLeft(s, 0, 4);
    }
    
    // 从s[start->end]里提取cnt个ip域结果放到ips中
    vector<string> getLeft(string& s, int start, int cnt) {
        int len = s.length();
        vector<string> ips;
        if (cnt == 1) {
            if (valid(s, start)) {
                ips.push_back(s.substr(start));
            }
            return ips;
        }
        
        for (int char_num = 1; char_num < 4; char_num++) {
            if (valid(s, start, char_num)) {
                string ip1 = s.substr(start, char_num) + ".";
                vector<string> subs = getLeft(s, start+char_num, cnt-1);
                for (int i = 0; i < subs.size(); i++) {
                    string ip2 = ip1 + subs[i];
                    ips.push_back(ip2);
                    //cout << "s=" << s << " cnt=" << cnt << " ip=" << ip2 << endl;
                }
            }
        }
        return ips;
    }
    
    // s[start.. start+char_num-1]的是否是有效的
    bool valid(string& s, int start, int char_num) {
        int len = s.length();
        switch(char_num) {
            case 1 : return start < len;
            case 2 : return start < len - 1 && s[start] != '0';
            case 3 : return valid3(s, start);
            default: return false;
        }
    }
    
    // [000-255]
    bool valid3(string&s, int start) {
        if (start < s.length() - 2) {
            int t = (s[start]-'0')*100 + (s[start+1]-'0')*10 + (s[start+2]-'0');
            if (t >= 100 && t <= 255) { return true; }
        }
        return false;
    }
    
    // s[start->end]是否有效
    bool valid(string&s , int start) {
        int len = s.length();
        int char_num = len - start;
        if (char_num > 3 || char_num < 1) { return false; }
        int ip_num = 0;
        while (start < len) {
            ip_num = ip_num * 10 + s[start++] - '0';
        }
        
        int min_ip = 0;
        if (char_num > 1) {
            min_ip = char_num == 2 ? 10 : 100;
        }
        if (ip_num >= min_ip && ip_num < 256) { return true; }
        return false;
    }
};