題目

93. 復原IP地址

我的思路

比較明顯，遞迴深搜處理比較方便。設定一個指標指向字串首，每次可以後移1,2,或3個位置。總共需要後移4次，且剛好滑過整個字串。同時要保證每次劃過的數字在0-255之間即可。

我的實現

class Solution {
public:
    vector<string> result;
    string ss;
    void process(const int length,int pos,int id,string str)
    {
        //cout<<"check"<<pos<<"\t"<<id<<"\t"<<str<<endl;
 

        if(pos>length)return;
        if(id>4&&pos==length){str.erase(str.end()-1);result.push_back(str);}
        if((5-id)*3<length-pos||(5-id)>length-pos)return;
        
            str = str + ss[pos++];
            process(length,pos,id+1,str+'.');
            if(ss[pos-1]=='
 
0')return;
            str = str + ss[pos++];
            
            process(length,pos,id+1,str+'.');
            str = str + ss[pos++];
            if(ss[pos-3]-'0'>2||(ss[pos-2]-'0'>5&&ss[pos-3]-'0'==2)||(ss[pos-2]-'0'==5&&ss[pos-3]-'0'==2&&ss[pos-1]-'0'>5))return;
            process(length,pos,id
 
+1,str+'.');
            
        
    }
    vector<string> restoreIpAddresses(string s) {
        
        ss = s;
        vector<string> a;
        if(s.size()<4||s.size()>12)return a;
        int pos = 0;
        int id = 1;
        string str;
        process(s.size(),pos,id,str);
        return result;

    }
};
/*
取
*/

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