leetcode-Restore IP Addresses-ZZ

http://www.cnblogs.com/remlostime/archive/2012/11/14/2770072.html

 class Solution {
 private:
     vector<string> ret;
     int pos[];
 public:
     bool check(string &s, int beg, int end)
     {
         string ip(s, beg, end - beg + );
         if (ip.size() == )
             return "" <= ip && ip <= "";
         else if (ip.size() == )
             return "" <= ip && ip <= "";
         else if (ip.size() == )
             return "" <= ip && ip <= "";
         else
             return false;
     }
 
     void dfs(int dep, int maxDep, string &s, int start)
     {
         if (dep == maxDep)
         {
             if (start == s.size())
             {
                 int beg = ;
                 string addr;
                 for(int i = ; i < maxDep; i++)
                 {
                     string ip(s, beg, pos[i] - beg + );
                     beg = pos[i] + ;
                     addr += i ==  ? ip : "." + ip;
                 }
                 ret.push_back(addr);
             }
             return;
         }
 
         for(int i = start; i < s.size(); i++)
             if (check(s, start, i))
             {
                 pos[dep] = i;
                 dfs(dep + , maxDep, s, i + );
             }
     }
 
     vector<string> restoreIpAddresses(string s) {
         // Start typing your C/C++ solution below
         // DO NOT write int main() function
         ret.clear();
         dfs(, , s, );
         return ret;
     }
 };

http://yucoding.blogspot.com/2013/04/leetcode-question-83-restore-ip.html

Restore IP Addresses:

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)

Analysis:
This problem can be viewed as a DP problem. There needed 3 dots to divide the string, and make sure the IP address is valid:  less than or equal to 255, greater or equal to 0, and note that, "0X" or "00X" is not valid.
For the DP, the length of each part is from 1 to 3. We use a vector<string> to store each part, and cut the string every time. Details see the code.

Note that "atoi" is for c-string, <string> need to convert to cstring by str.c_str();

Code(Updated 201309):

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class Solution { public:     bool valid(string s){         if (s.size()==3 && (atoi(s.c_str())>255 || atoi(s.c_str())==0)){return false;}         if (s.size()==3 && s[0]=='0'){return false;}         if (s.size()==2 && atoi(s.c_str())==0){return false;}         if (s.size()==2 && s[0]=='0'){return false;}         return true;     }       void getRes(string s, string r, vector<string> &res, int k){         if (k==0){             if (s.empty()){res.push_back(r);}             return;         }else{             for (int i=1;i<=3;i++){                 if (s.size()>=i && valid(s.substr(0,i))){                     if (k==1){getRes(s.substr(i),r+s.substr(0,i),res,k-1);}                     else{getRes(s.substr(i),r+s.substr(0,i)+".",res,k-1);}                 }             }         }     }       vector<string> restoreIpAddresses(string s) {         // Start typing your C/C++ solution below         // DO NOT write int main() function         vector<string> res;         getRes(s,"",res,4);         return res;     } };

Code(old version):

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class Solution { public:       void dp(string s,vector<string> &cur ,vector<string> &res){                       if (cur.size()==3){ // if there are 4 parts in the original string             cur.push_back(s); //all 4 parts and check if valid             bool r = true;             for (int i=0;i<4;i++){                 if (atoi(cur[i].c_str())>255){  //check value                     r = false;                     break;                 }                 if ((cur[i].size()>1 && cur[i][0]=='0')){ //check "0X" "00X" and "0XX" cases                     r =false;                     break;                 }             }                    if (r){                 res.push_back(cur[0]+"."+cur[1]+"."+cur[2]+"."+cur[3]);             }             cur.pop_back();                       }else{             for (int i=0;i<3;i++){                 if (s.size()>i+1){                     cur.push_back(s.substr(0,i+1));                     dp(s.substr(i+1,(s.size()-i-1)),cur,res);                     cur.pop_back();                 }             }         }               }         vector<string> restoreIpAddresses(string s) {         // Start typing your C/C++ solution below         // DO NOT write int main() function         vector<string> res,cur;         if (s.size()>12 || s.size()<4 ){return res;}         dp(s,cur,res); // cur stores the current separation         return res;     } };