### leetcode Question 83: 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):

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);}
}
}
}
}

// 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):

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();
}
}
}

}