最好的解决方案是这样的一句话:

String hex=DatatypeConverter.printHexBinary(byte[] b);

正如这里提到的

    byte[] bytes = {-1, 0, 1, 2, 3 };
    StringBuilder sb = new StringBuilder();
    for (byte b : bytes) {
        sb.append(String.format("%02X ", b));
    }
    System.out.println(sb.toString());
    // prints "FF 00 01 02 03 "

另请参阅

java.util.Formatter语法 %(旗帜)(宽度)转换 标记'0' -结果将被填充为零 宽度2 转换'X' -结果被格式化为十六进制整数，大写

看看问题的文本，也有可能是这样要求的:

    String[] arr = {"-1", "0", "10", "20" };
    for (int i = 0; i < arr.length; i++) {
        arr[i] = String.format("%02x", Byte.parseByte(arr[i]));
    }
    System.out.println(java.util.Arrays.toString(arr));
    // prints "[ff, 00, 0a, 14]"

这里的几个答案使用Integer.toHexString(int);这是可行的，但有一些注意事项。由于形参是int型，因此对byte参数执行扩大原语转换，这涉及到符号扩展。

    byte b = -1;
    System.out.println(Integer.toHexString(b));
    // prints "ffffffff"

在Java中有符号的8位字节被符号扩展为32位整型。为了有效地撤销这个符号扩展，可以使用0xFF来屏蔽字节。

    byte b = -1;
    System.out.println(Integer.toHexString(b & 0xFF));
    // prints "ff"

使用toHexString的另一个问题是它不会用零填充:

    byte b = 10;
    System.out.println(Integer.toHexString(b & 0xFF));
    // prints "a"

这两个因素结合起来应该形成字符串。格式解决方案更佳。

参考文献

整型类型和值 对于字节，从-128到127，包括 JLS 5.1.2扩大原语转换

如果存在性能问题，创建(并销毁)一堆String实例并不是一个好方法。

请忽略那些冗长(重复)参数检查语句(if)。那是为了(另一个)教育目的。

完整的maven项目:http://jinahya.googlecode.com/svn/trunk/com.googlecode.jinahya/hex-codec/

编码…

/**
 * Encodes a single nibble.
 *
 * @param decoded the nibble to encode.
 *
 * @return the encoded half octet.
 */
protected static int encodeHalf(final int decoded) {

    switch (decoded) {
        case 0x00:
        case 0x01:
        case 0x02:
        case 0x03:
        case 0x04:
        case 0x05:
        case 0x06:
        case 0x07:
        case 0x08:
        case 0x09:
            return decoded + 0x30; // 0x30('0') - 0x39('9')
        case 0x0A:
        case 0x0B:
        case 0x0C:
        case 0x0D:
        case 0x0E:
        case 0x0F:
            return decoded + 0x57; // 0x41('a') - 0x46('f')
        default:
            throw new IllegalArgumentException("illegal half: " + decoded);
    }
}


/**
 * Encodes a single octet into two nibbles.
 *
 * @param decoded the octet to encode.
 * @param encoded the array to which each encoded nibbles are written.
 * @param offset the offset in the array.
 */
protected static void encodeSingle(final int decoded, final byte[] encoded,
                                   final int offset) {

    if (encoded == null) {
        throw new IllegalArgumentException("null encoded");
    }

    if (encoded.length < 2) {
        // not required
        throw new IllegalArgumentException(
            "encoded.length(" + encoded.length + ") < 2");
    }

    if (offset < 0) {
        throw new IllegalArgumentException("offset(" + offset + ") < 0");
    }

    if (offset >= encoded.length - 1) {
        throw new IllegalArgumentException(
            "offset(" + offset + ") >= encoded.length(" + encoded.length
            + ") - 1");
    }

    encoded[offset] = (byte) encodeHalf((decoded >> 4) & 0x0F);
    encoded[offset + 1] = (byte) encodeHalf(decoded & 0x0F);
}


/**
 * Decodes given sequence of octets into a sequence of nibbles.
 *
 * @param decoded the octets to encode
 *
 * @return the encoded nibbles.
 */
protected static byte[] encodeMultiple(final byte[] decoded) {

    if (decoded == null) {
        throw new IllegalArgumentException("null decoded");
    }

    final byte[] encoded = new byte[decoded.length << 1];

    int offset = 0;
    for (int i = 0; i < decoded.length; i++) {
        encodeSingle(decoded[i], encoded, offset);
        offset += 2;
    }

    return encoded;
}


/**
 * Encodes given sequence of octets into a sequence of nibbles.
 *
 * @param decoded the octets to encode.
 *
 * @return the encoded nibbles.
 */
public byte[] encode(final byte[] decoded) {

    return encodeMultiple(decoded);
}

解码…

/**
 * Decodes a single nibble.
 *
 * @param encoded the nibble to decode.
 *
 * @return the decoded half octet.
 */
protected static int decodeHalf(final int encoded) {

    switch (encoded) {
        case 0x30: // '0'
        case 0x31: // '1'
        case 0x32: // '2'
        case 0x33: // '3'
        case 0x34: // '4'
        case 0x35: // '5'
        case 0x36: // '6'
        case 0x37: // '7'
        case 0x38: // '8'
        case 0x39: // '9'
            return encoded - 0x30;
        case 0x41: // 'A'
        case 0x42: // 'B'
        case 0x43: // 'C'
        case 0x44: // 'D'
        case 0x45: // 'E'
        case 0x46: // 'F'
            return encoded - 0x37;
        case 0x61: // 'a'
        case 0x62: // 'b'
        case 0x63: // 'c'
        case 0x64: // 'd'
        case 0x65: // 'e'
        case 0x66: // 'f'
            return encoded - 0x57;
        default:
            throw new IllegalArgumentException("illegal half: " + encoded);
    }
}


/**
 * Decodes two nibbles into a single octet.
 *
 * @param encoded the nibble array.
 * @param offset the offset in the array.
 *
 * @return decoded octet.
 */
protected static int decodeSingle(final byte[] encoded, final int offset) {

    if (encoded == null) {
        throw new IllegalArgumentException("null encoded");
    }

    if (encoded.length < 2) {
        // not required
        throw new IllegalArgumentException(
            "encoded.length(" + encoded.length + ") < 2");
    }

    if (offset < 0) {
        throw new IllegalArgumentException("offset(" + offset + ") < 0");
    }

    if (offset >= encoded.length - 1) {
        throw new IllegalArgumentException(
            "offset(" + offset + ") >= encoded.length(" + encoded.length
            + ") - 1");
    }

    return (decodeHalf(encoded[offset]) << 4)
           | decodeHalf(encoded[offset + 1]);
}


/**
 * Encodes given sequence of nibbles into a sequence of octets.
 *
 * @param encoded the nibbles to decode.
 *
 * @return the encoded octets.
 */
protected static byte[] decodeMultiple(final byte[] encoded) {

    if (encoded == null) {
        throw new IllegalArgumentException("null encoded");
    }

    if ((encoded.length & 0x01) == 0x01) {
        throw new IllegalArgumentException(
            "encoded.length(" + encoded.length + ") is not even");
    }

    final byte[] decoded = new byte[encoded.length >> 1];

    int offset = 0;
    for (int i = 0; i < decoded.length; i++) {
        decoded[i] = (byte) decodeSingle(encoded, offset);
        offset += 2;
    }

    return decoded;
}


/**
 * Decodes given sequence of nibbles into a sequence of octets.
 *
 * @param encoded the nibbles to decode.
 *
 * @return the decoded octets.
 */
public byte[] decode(final byte[] encoded) {

    return decodeMultiple(encoded);
}

Java 17:引入Java .util. hexformat

Java 17提供了一个实用程序，可以将字节数组和数字转换为对应的十六进制数。假设我们有一个“Hello World”的MD5摘要作为一个字节数组:

var md5 = MessageDigest.getInstance("md5");
md5.update("Hello world".getBytes(UTF_8));

var digest = md5.digest();

现在我们可以使用HexFormat.of().formatHex(byte[])方法将给定的byte[]转换为它的十六进制形式:

jshell> HexFormat.of().formatHex(digest)
$7 ==> "3e25960a79dbc69b674cd4ec67a72c62"

withUpperCase()方法返回前面输出的大写版本:

jshell> HexFormat.of().withUpperCase().formatHex(digest)
$8 ==> "3E25960A79DBC69B674CD4EC67A72C62"

如果你想要一个恒定宽度的十六进制表示，即0A而不是a，这样你就可以明确地恢复字节，尝试format():

StringBuilder result = new StringBuilder();
for (byte bb : byteArray) {
    result.append(String.format("%02X", bb));
}
return result.toString();

其他人已经报道了一般情况。但是如果你有一个已知形式的字节数组，例如MAC地址，那么你可以:

byte[] mac = { (byte)0x00, (byte)0x00, (byte)0x00, (byte)0x00, (byte)0x00 };

String str = String.format("%02X:%02X:%02X:%02X:%02X:%02X",
                           mac[0], mac[1], mac[2], mac[3], mac[4], mac[5]);