class Solution {
public:
    vector<string> res;
    vector<string> restoreIpAddresses(string s) {
        DFS(res,s,"",0,0);
        return res;
    }
    void DFS(vector<string>& res,string s,string t,int count,int k)   //count表示段数，k表示s遍历的位置
    {
        if(count==4&&k==s.size())
        {
            res.push_back(t);
            return;
        }
        else if(count>3)
            return;
        string tmp1=t;
        if(s[k]=='0')        //当前元素为0时，即为一段
        {
            tmp1.push_back(s[k]);
            if(count!=3)
                tmp1.push_back('.');
            DFS(res,s,tmp1,count+1,k+1);
        }
        else
        {
            int sum=0;
            for(int i=k;i<s.size();i++)
            {
                sum=sum*10+(s[i]-'0');
                if(sum<=255)
                {
                    string tmp2=tmp1;              //tmp1为未打点，tmp2已打点成为一段
                    tmp2.push_back(s[i]);
                    tmp1.push_back(s[i]);
                    if(count!=3)                  //第四段即count==3时，尾部不打点
                        tmp2.push_back('.');
                    DFS(res,s,tmp2,count+1,i+1);
                }
                else 
                    break;
            }
        }
    }
};