判断IP是否合法

现在不需要自己写正则表达式去匹配IP是否合法，只需要调用一个简单的函数就可以解决问题。

String ip = "xxx.xxx.xxx.xxx";boolean isIpString = InetAddress.isNumeric(ip);

返回true，表示合法；返回false，表示不是合法的IP。

/**     * Returns true if the string is a valid numeric IPv4 or IPv6 address (such as "192.168.0.1").     * This copes with all forms of address that Java supports, detailed in the {@link InetAddress}     * class documentation.     *     * @hide used by frameworks/base to ensure that a getAllByName won't cause a DNS lookup.     */    public static boolean isNumeric(String address) {        InetAddress inetAddress = parseNumericAddressNoThrow(address);        return inetAddress != null && disallowDeprecatedFormats(address, inetAddress) != null;    } private static InetAddress parseNumericAddressNoThrow(String address) {        // Accept IPv6 addresses (only) in square brackets for compatibility.        if (address.startsWith("[") && address.endsWith("]") && address.indexOf(':') != -1) {            address = address.substring(1, address.length() - 1);        }        StructAddrinfo hints = new StructAddrinfo();        hints.ai_flags = AI_NUMERICHOST;        InetAddress[] addresses = null;        try {            addresses = Libcore.os.getaddrinfo(address, hints);        } catch (GaiException ignored) {        }        return (addresses != null) ? addresses[0] : null;    } private static InetAddress disallowDeprecatedFormats(String address, InetAddress inetAddress) {        // Only IPv4 addresses are problematic.        if (!(inetAddress instanceof Inet4Address) || address.indexOf(':') != -1) {            return inetAddress;        }        // If inet_pton(3) can't parse it, it must have been a deprecated format.        // We need to return inet_pton(3)'s result to ensure that numbers assumed to be octal        // by getaddrinfo(3) are reinterpreted by inet_pton(3) as decimal.        return Libcore.os.inet_pton(AF_INET, address);    }