Leetcode 468 Validate IP Address

最新推荐文章于 2023-07-29 11:50:48 发布

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最新推荐文章于 2023-07-29 11:50:48 发布

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本文介绍了一个用于验证IPv4和IPv6地址有效性的Java方法。通过详细解析IPv4和IPv6地址的构成规则，提供了两个静态方法来分别验证这两种类型的IP地址。文章深入探讨了如何正确处理IP地址中的各种特殊情况。

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Write a function to check whether an 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, leading zeros in the IPv4 is invalid. For example, the address 172.16.254.01 is invalid.

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 valid 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, extra leading zeros in the IPv6 is also invalid. For example, the address 02001:0db8:85a3:0000:0000:8a2e:0370:7334 is invalid.

Note: You may assume there is no extra space or 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.

在正则表达式中 .可以表示任意字符

所有字符都作为分隔符当然不会有任何结果
\\.实际上被转义为两次，\\在java中被转换为一个'\'字符，然后'\.'被传给正则，\.表示对点字符进行转义，使.就表示字符'.',而不使用它在正则中的特殊意义

一个string的问题。值得注意的地方是如何判断是否符合IP的要求。

因为要判断俩，用if就比较繁杂，考虑分开用static method写吧

public class Solution {
   public String validIPAddress(String IP) {
	if(isValidIPv4(IP)) return "IPv4";
	else if(isValidIPv6(IP)) return "IPv6";
	else return "Neither";
}

public boolean isValidIPv4(String ip) {
	if(ip.length()<7) return false;
	if(ip.charAt(0)=='.') return false;
	if(ip.charAt(ip.length()-1)=='.') return false;
	String[] tokens = ip.split("\\.");
	if(tokens.length!=4) return false;
	for(String token:tokens) {
		if(!isValidIPv4Token(token)) return false;
	}
	return true;
}
public boolean isValidIPv4Token(String token) {
	if(token.startsWith("0") && token.length()>1) return false;
	try {
		int parsedInt = Integer.parseInt(token);
		if(parsedInt<0 || parsedInt>255) return false;
		if(parsedInt==0 && token.charAt(0)!='0') return false;
	} catch(NumberFormatException nfe) {
		return false;
	}
	return true;
}
	
public boolean isValidIPv6(String ip) {
	if(ip.length()<15) return false;
	if(ip.charAt(0)==':') return false;
	if(ip.charAt(ip.length()-1)==':') return false;
	String[] tokens = ip.split(":");
	if(tokens.length!=8) return false;
	for(String token: tokens) {
		if(!isValidIPv6Token(token)) return false;
	}
	return true;
}
public boolean isValidIPv6Token(String token) {
	if(token.length()>4 || token.length()==0) return false;
	char[] chars = token.toCharArray();
	for(char c:chars) {//在IPv6里面 有大写字母小写字母和数字三种情况 要对每一个char都进行检查
		boolean isDigit = c>=48 && c<=57;
		boolean isUppercaseAF = c>=65 && c<=70;
		boolean isLowerCaseAF = c>=97 && c<=102;
		if(!(isDigit || isUppercaseAF || isLowerCaseAF)) 
			return false;
	}
	return true;
}
}