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)

1.我的答案 回溯

class Solution {
public:
void split(vector<string>&res, string s, vector<string>& vec, int num){
int len = s.size();
if(num == 0 && len > 0 && len < 4){
if(s.size()>1 && s[0] == '0')
return;
else{
int temp = atoi(s.c_str());
if(temp <=255 && temp >=0){
string sres = vec[0]+"."+vec[1]+"."+vec[2]+"."+s;
res.push_back(sres);
return;
}
}
}
int i = 1;
while(i <=3 && i < len){
string stemp = s.substr(0,i);
if(stemp.size()>1 && stemp[0] == '0'){//用来判断是否有“010”这种情况出现，出现就删掉
++i;
continue;
}
else{
int sint = atoi(stemp.c_str());
if(sint <= 255 && sint >=0){
vec.push_back(stemp);
string str2 = s.substr(i,len-i);
split(res,str2,vec,num-1);
vec.pop_back();
}
}
++i;
}
return;
}

vector<string> restoreIpAddresses(string s) {
vector<string>res,vec;
int num = 3;
int len = s.size();
if(len < 4 || len > 12)
return res;
split(res,s,vec,num);
return res;
}
};

2.别人的答案，时间一样，就是比我的要整洁

class Solution {
public:
vector<string> restoreIpAddresses(string s) {
vector<string> result;
string ip;
dfs(s,0,0,ip,result); //paras:string s,start index of s,step(from0-3),intermediate ip,final result
return result;
}
void dfs(string s,int start,int step,string ip,vector<string>& result){
if(start==s.size()&&step==4){
ip.erase(ip.end()-1); //remove the last '.' from the last decimal number
result.push_back(ip);
return;
}
if(s.size()-start>(4-step)*3) return;
if(s.size()-start<(4-step)) return;
int num=0;
for(int i=start;i<start+3;i++){
num=num*10+(s[i]-'0');
if(num<=255){
ip+=s[i];
dfs(s,i+1,step+1,ip+'.',result);
}
if(num==0) break;
}
}
};