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)

class Solution {
public:
vector<string> ans;
vector<string> restoreIpAddresses(string s) {
ans.clear();
vector<string> path;
dfs(s,path,0);
return ans;
}
bool isOk(string s, int start, int end){
int sum = 0;
int b = 1;
//比如0XX 0X 都不允许
if (s[start] == '0' && end - start > 1){
return false;
}
for(int i = end -1; i >= start; i--){
sum +=  (s[i] - '0')*b;
b *= 10;
}
if (sum >= 0 && sum < 256){
return true;
}
return false;
}
void dfs(string s, vector<string> & path, int pos){
if (path.size() == 4){
if (pos >= s.size()){
string one = path[0];
for(int i = 1; i < path.size(); i++){
one += ".";
one += path[i];
}
ans.push_back(one);
return;
}
}
for(int i = 1; i < 4; i++){
if (pos + i <= s.size() && isOk(s,pos,pos+i)){
//剩下的长度过长
int left = s.size() - pos - i;
if (path.size() + left/3.0 > 4){
continue;
}
path.resize(path.size() +1);
path[path.size() -1] = s.substr(pos,i);
dfs(s,path,pos+i);
path.resize(path.size() -1);
}
}
}
};

using namespace std;
int main(int argc, char *argv[]) {
string s = "010010";
Solution sol;
vector<string> ans = sol.restoreIpAddresses(s);
for(int i = 0; i < ans.size(); i++){
cout << ans[i] << endl;
}
}