93. 复原 IP 地址

自己做

解:回溯法
cpp
class Solution {
private:
vector<string> res;
public:
void tryIpAddresses(string &s, int index, int i, string &ip){ //index代表s的下标索引,i代表是第几段IP地址(分四段),ip即是保存的结果
if((int)s.size() - index > (4 - i) * 3 ) //取不完完整的IP地址,比如最多取12位(3 * 4),但是字符串却有20位,这时候就取不完了,必然组不成IP地址
return;
if(ip.size() - 1 == s.size() + 3 && i == 4){ //保存结果【已经取完四段IP地址】
ip.pop_back(); //弹出'.'
res.push_back(ip);
ip.push_back('.');
return;
}
//尝试取一位
ip.push_back(s[index]);
ip.push_back('.');
tryIpAddresses(s, index + 1, i + 1, ip);
ip.pop_back();
ip.pop_back();
//尝试取两位
if(index + 1 < (int)s.size() && s[index] > '0'){
ip.push_back(s[index]);
ip.push_back(s[index + 1]);
ip.push_back('.');
tryIpAddresses(s, index + 2, i + 1, ip);
ip.pop_back();
ip.pop_back();
ip.pop_back();
}
//尝试取三位
if(index + 2 < (int)s.size() &&
(s[index] == '1' ||
s[index] == '2' && s[index + 1] == '5' && s[index + 2] <= '5' ||
s[index] == '2' && s[index + 1] < '5')){
ip.push_back(s[index]);
ip.push_back(s[index + 1]);
ip.push_back(s[index + 2]);
ip.push_back('.');
tryIpAddresses(s, index + 3, i + 1, ip);
ip.pop_back();
ip.pop_back();
ip.pop_back();
ip.pop_back();
}
}
vector<string> restoreIpAddresses(string s) {
string str;
tryIpAddresses(s, 0, 0, str);
return res;
}
};
