[LeetCode] Validate IP Address 驗證IP地址

 

In this problem, your job to write a function to check whether a input string is a valid IPv4 address or IPv6 address or neither.

IPv4 addresses are canonically represented in dot-decimal notation, which consists of four decimal numbers, each ranging from 0 to 255, separated by dots ("."), e.g.,172.16.254.1;

Besides, you need to keep in mind that leading zeros in the IPv4 is illegal. For example, the address 172.16.254.01 is illegal.

IPv6 addresses are represented as eight groups of four hexadecimal digits, each group representing 16 bits. The groups are separated by colons (":"). For example, the address 2001:0db8:85a3:0000:0000:8a2e:0370:7334 is a legal one. Also, we could omit some leading zeros among four hexadecimal digits and some low-case characters in the address to upper-case ones, so 2001:db8:85a3:0:0:8A2E:0370:7334 is also a valid IPv6 address(Omit leading zeros and using upper cases).

However, we don't replace a consecutive group of zero value with a single empty group using two consecutive colons (::) to pursue simplicity. For example, 2001:0db8:85a3::8A2E:0370:7334 is an invalid IPv6 address.

Besides, you need to keep in mind that extra leading zeros in the IPv6 is also illegal. For example, the address 02001:0db8:85a3:0000:0000:8a2e:0370:7334 is also illegal.

Note: You could assume there is no extra space in the test cases and there may some special characters in the input string.

Example 1:

Input: "172.16.254.1"

Output: "IPv4"

Explanation: This is a valid IPv4 address, return "IPv4".

 

Example 2:

Input: "2001:0db8:85a3:0:0:8A2E:0370:7334"

Output: "IPv6"

Explanation: This is a valid IPv6 address, return "IPv6".

 

Example 3:

Input: "256.256.256.256"

Output: "Neither"

Explanation: This is neither a IPv4 address nor a IPv6 address.

 

這道題讓咱們驗證兩種IP地址，LeetCode以前有一道關於IPv4的題Restore IP Addresses，給咱們了一個字符串，讓咱們經過在中間加點來找出全部正確的IP地址，這道題給了咱們中間加點或者冒號的字符串，讓咱們驗證其是不是正確的IPv4或者IPv6，感受要稍稍複雜一些。那麼咱們只有分別來驗證了，那麼咱們怎麼樣能快速的區別是IPv4或者IPv6呢，固然是經過中間的點或者冒號啦，因此咱們首先在字符串中找冒號(固然你想找點也能夠)，若是字符串中沒有冒號，那麼咱們來驗證其是不是IPv4，若是有冒號，咱們就來驗證其是不是IPv6.

首先對於IPv4，咱們使用getline函數來截取兩個點之間的字符串，咱們還須要一個計數器cnt來記錄咱們已經截取了多少段，若是cnt大於4了，說明超過了4段，說明是否是正確的地址。若是取出的字符串爲空，說明兩個點連在一塊兒了，也不對。再有就是若是字符串長度大於1，且第一個字符是0，也不對。因爲IPv4的地址在0到255之間，因此若是字符串長度大於3，也不正確。下面咱們檢查每個字符，若是有不是數字的字符，返回Neither。最後咱們再把字符串轉爲數字，若是不在0到255之間就是非法的。最後的最後，咱們要保證cnt正好爲4，並且最後一個字符不能是點，通通知足以上條件纔是正確的IPv4地址。

而後對於IPv6，咱們也使用getline函數來截取兩個冒號之間的字符串，咱們一樣須要計數器cnt來記錄咱們已經截取了多少段，若是cnt大於8了，說明超過了8段，說明是否是正確的地址。若是取出的字符串爲空，說明兩個冒號連在一塊兒了，也不對。面咱們檢查每個字符，正確的字符應該是0到9之間的數字，或者a到f，或A到F之間的字符，若是出現了其餘字符，返回Neither。最後的最後，咱們要保證cnt正好爲8，並且最後一個字符不能是冒號，通通知足以上條件纔是正確的IPv6地址。

 

class Solution {
public:
    string validIPAddress(string IP) {
        istringstream is(IP);
        string t = "";
        int cnt = 0;
        if (IP.find(':') == string::npos) { // Check IPv4
            while (getline(is, t, '.')) {
                ++cnt;
                if (cnt > 4 || t.empty() || (t.size() > 1 && t[0] == '0') || t.size() > 3) return "Neither";
                for (char c : t) {
                    if (c < '0' || c > '9') return "Neither";
                }
                int val = stoi(t);
                if (val < 0 || val > 255) return "Neither";
            }
            return (cnt == 4 && IP.back() != '.') ? "IPv4" : "Neither";
        } else { // Check IPv6
            while (getline(is, t, ':')) {
                ++cnt;
                if (cnt > 8 || t.empty() || t.size() > 4) return "Neither";
                for (char c : t) {
                    if (!(c >= '0' && c <= '9') && !(c >= 'a' && c <= 'f') && !(c >= 'A' && c <= 'F')) return "Neither";
                }
            }
            return (cnt == 8 && IP.back() != ':') ? "IPv6" : "Neither";
        }
    }
};

 

相似題目：

Restore IP Addresses

 

參考資料：

https://discuss.leetcode.com/topic/71572/java-solution

https://discuss.leetcode.com/topic/71418/short-regexp-solution/5

 

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