Leetcode 468. Validate IP Address 验证IP地址 解题报告

1 解题思想

这道题就是给一个字符串，验证其是否是IPV4或者是IPV6的格式。
题目思想就是要么正则，要么不用正则，自己加规则。。。这道题我没做，discuss上找的改的，这里声明下，因为觉得觉得有点重复劳动的工作。

这里解释下规则：
IPV4：四个数，三个点分隔，每个数的范围是0~255，除了0外不允许0开头
IPV6：完全形态是8个16进制填满的四位数（四位全都有数，可以填充0），7个：分隔，每个数0开头的地方可以省略（也可以不），有连续的一组全0也可以省，可以大小字母写。。。

2 原题

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.

3 AC 解

public class Solution {    //https://discuss.leetcode.com/topic/71434/java-simple-solution-without-regex/3    public String validIPAddress(String IP) {        if(IP == null || IP.isEmpty()) return "Neither";        if(IP.contains(":")){            return (isV6(IP)?"IPv6":"Neither");        } else {            return (isV4(IP)?"IPv4":"Neither");        }    }    private boolean isV6(String ip){        int dotCount = 0;        for(int i = 0;i<ip.length();i++) if(ip.charAt(i) == ':') dotCount++;        if(dotCount != 7) return false;        String gr[]= ip.split(":");        if(gr.length != 8) return false;        for(String g:gr){            if(g.isEmpty()) return false;            if(g.length()> 4){                return false;            }            for(int i = 0;i<g.length();i++){                char ch = g.charAt(i);                if(!((ch >= '0' && ch <= '9') || (ch>='a' && ch<='f') || (ch>='A' &&ch<='F'))) return false;            }        }        return true;    }    private boolean isV4(String ip){        int dotCount = 0;        for(int i = 0;i<ip.length();i++) if(ip.charAt(i) == '.') dotCount++;        if(dotCount != 3) return false;        String gr[] = ip.split("\\.");        if(gr.length != 4) return false;        for(String g: gr){            if(g.isEmpty()) return false;            try{                int val =  Integer.parseInt(g);                //检查是否0开头，这里取巧了一些                if( (val+"").equals(g) == false ) return false;                if(val > 255 || val <0 ) return false;                if(g.charAt(0) == '0' && (val != 0 || g.length() != 1) ) return false;            }catch(NumberFormatException ex){                return false;            }        }        return true;    }}