93. Restore IP Addresses(dfs)

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)

思路：IP address的规则：一共四段；每段的值不能超过255；不能以0开头，但可以在一段中只有数字0

class Solution {
public:
    vector<string> restoreIpAddresses(string s) {
        dfs(s, "", );
        return ret;
    }

    void dfs(string s, string currentIP, int depth){ //depth表示第几个section
        if(depth == ){
            if(check(s))
                ret.push_back(currentIP+s);
            return;
        }

        int len = s.length();
        if(len < -depth){ //剩余string长度过短
            return;
        }

        string s1, s2;
        //check if we can assign 3 digits in the section
        if(len >  ){
            s1 = s.substr(,);

            if(check(s1)){
                s2 = s.substr();
                dfs(s2,currentIP+s1+".", depth+);
            }
        }

        //check if we can assign 2 digits in the section
        if(len >  ){
            s1 = s.substr(,);
            if(check(s1)){
                s2 = s.substr();
                dfs(s2,currentIP+s1+".", depth+);
            }
        }

        //assign 1 digits in the section
        s2 = s.substr();
        dfs(s2,currentIP+s[]+".", depth+);
    }

    bool check(string section){
        int len = section.length();
        if(len ==  || len > ) return false;
        int value = stoi(section);
        if(len==){
            if(section[]!='' && value <= ) return true;
            else return false;
        }
        else if(len==){
            if(section[]=='') return false;
            else return true;
        }
        return true;
    }
private:
    vector<string> ret;
};

当然也能用循环，每两个section之间的分割用一个for循环遍历分割的位置，一共是三重for循环。