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)
思路：
DFS。只需增加判断每个部分数字在0到255之间，并且第一位数不能是0.
另外代码中用到string转int的方法，先利用c_str()转成C string，再用atoi()与atof()。
代码：
1 vector<string> result; 2 string tmp; 3 void search(int step, string s){ 4 int l = s.length(); 5 if(l > (5-step)*3 || l < (5-step)) 6 return; 7 if(step == 4){ 8 if(s[0] == '0' && s.length() > 1) 9 return ; 10 int t = atoi(s.c_str()); 11 if(t >= 0 && t <= 255){ 12 tmp = tmp + "." + s; 13 result.push_back(tmp); 14 } 15 return; 16 } 17 if(l > 3) 18 l = 3; 19 if(s[0] == '0') 20 l = 1; 21 for(int i = 1; i <= l; i++){ 22 string s1 = s.substr(0, i); 23 string s2 = s.substr(i); 24 int t = atoi(s1.c_str()); 25 if(t >= 0 && t <= 255){ 26 string pre = tmp; 27 if(tmp == ""){ 28 tmp = s1; 29 search(step+1, s2); 30 tmp = ""; 31 } 32 else{ 33 tmp = tmp + "." + s1; 34 search(step+1, s2); 35 tmp = pre; 36 } 37 } 38 } 39 } 40 vector<string> restoreIpAddresses(string s) { 41 // IMPORTANT: Please reset any member data you declared, as 42 // the same Solution instance will be reused for each test case. 43 result.clear(); 44 tmp = ""; 45 search(1, s); 46 return result; 47 }