题目

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思路

  • 思路一：暴力根据规则验证

    代码

    1. public String validIPAddress(String IP) {
    2.       String[] ipv4 = IP.split("\\.");
    3.       String[] ipv6 = IP.split(":");
    4.       if (ipv4.length == 4 && IP.charAt(0) != '.' && IP.charAt(IP.length() - 1) != '.') {
    5.           for (String s : ipv4) {
    6.               if (s.length() == 0 || s.length() >= 4 
    7.               || s.charAt(0) == '0' && s.length() > 1) return "Neither";
    8.               int num = 0;
    9.               for (int i = 0; i < s.length(); i++) {
    10.                   char ch = s.charAt(i);
    11.                   if (ch < '0' || ch > '9') return "Neither";
    12.                   num = num * 10 + ch - '0';
    13.               }
    14.               if (num >= 256) return "Neither";
    15.           }
    16.           return "IPv4";
    17.       }
    18.       if (ipv6.length == 8 && IP.length() >= 15 && IP.charAt(0) != ':' 
    19.       && IP.charAt(IP.length() - 1) != ':') {
    20.           for (String s : ipv6) {
    21.               if (s.length() == 0 || s.length() > 4) return "Neither";
    22.               for (int i = 0; i < s.length(); i++) {
    23.                   char ch = s.charAt(i);
    24.                   if (!(ch >= '0'  && ch <= '9' || ch >= 'A' && ch <= 'F' 
    25.                   || ch >= 'a' && ch <= 'f')) return "Neither";
    26.               }
    27.           }
    28.           return "IPv6";
    29.       }
    30.       return "Neither";
    31.   }
    验证IP地址