Solution
There are no special optimization or tricks to solve this question. The solution is rather straightforward and only contains string manipulation. Here is a simple and elaborate solution by yangluphil:
class Solution {
public:
const string validIPv6Chars = "0123456789abcdefABCDEF";
bool isValidIPv4Block(string& block) {
int num = 0;
if (block.size() > 0 && block.size() <= 3) {
for (int i = 0; i < block.size(); i++) {
char c = block[i];
// special case: if c is a leading zero and there are characters left
if (!isalnum(c) || (i == 0 && c == '0' && block.size() > 1))
return false;
else {
num *= 10;
num += c - '0';
}
}
return num <= 255;
}
return false;
}
bool isValidIPv6Block(string& block) {
if (block.size() > 0 && block.size() <= 4) {
for (int i = 0; i < block.size(); i++) {
char c = block[i];
if (validIPv6Chars.find(c) == string::npos)
return false;
}
return true;
}
return false;
}
string validIPAddress(string IP) {
string ans[3] = {"IPv4", "IPv6", "Neither"};
stringstream ss(IP);
string block;
// ipv4 candidate
if (IP.substr(0, 4).find('.') != string::npos) {
for (int i = 0; i < 4; i++) {
if (!getline(ss, block, '.') || !isValidIPv4Block(block))
return ans[2];
}
return ss.eof() ? ans[0] : ans[2];
}
// ipv6 candidate
else if (IP.substr(0, 5).find(':') != string::npos) {
for (int i = 0; i < 8; i++) {
if (!getline(ss, block, ':') || !isValidIPv6Block(block))
return ans[2];
}
return ss.eof() ? ans[1] : ans[2];
}
return ans[2];
}
};