题目描述:
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)
题目解析:
给一个字符串s,求这个字符串所能表示的所有的ip地址。采用深度优先遍历,如果字符串转换的个数为3且数字范围在0-255(合法,需要先将字符串转换为整型),就放进vector中。
class Solution {
public:
vector<string> restoreIpAddresses(string s) {
vector<string> res;
string tmp;
DFS(res,tmp,s,0);
return res;
}
void DFS(vector<string> &res,string tmp,string s,int count){
if(count == 3 && isValid(s)) //合法并且个数为3
{
res.push_back(tmp+s);
return;
}
for(int i = 1;i < 4 && i < s.size(); ++i){
string sub = s.substr(0,i);
if(isValid(sub))
DFS(res,tmp+sub+'.',s.substr(i),count+1);
}
}
bool isValid(string &s){
stringstream ss(s);
int num;
ss >> num; //将字符串转化为整型
//num = atoi(s.c_str());
if(s.size() > 1)
return s[0] != '0' && num >= 0 && num <= 255;
return num >= 0 && num <= 255;
}
};