在存储查找表的很小代价下，这个实现非常简单和快速。

 private static final char[] BYTE2HEX=(
    "000102030405060708090A0B0C0D0E0F"+
    "101112131415161718191A1B1C1D1E1F"+
    "202122232425262728292A2B2C2D2E2F"+
    "303132333435363738393A3B3C3D3E3F"+
    "404142434445464748494A4B4C4D4E4F"+
    "505152535455565758595A5B5C5D5E5F"+
    "606162636465666768696A6B6C6D6E6F"+
    "707172737475767778797A7B7C7D7E7F"+
    "808182838485868788898A8B8C8D8E8F"+
    "909192939495969798999A9B9C9D9E9F"+
    "A0A1A2A3A4A5A6A7A8A9AAABACADAEAF"+
    "B0B1B2B3B4B5B6B7B8B9BABBBCBDBEBF"+
    "C0C1C2C3C4C5C6C7C8C9CACBCCCDCECF"+
    "D0D1D2D3D4D5D6D7D8D9DADBDCDDDEDF"+
    "E0E1E2E3E4E5E6E7E8E9EAEBECEDEEEF"+
    "F0F1F2F3F4F5F6F7F8F9FAFBFCFDFEFF").toCharArray();
   ; 

  public static String getHexString(byte[] bytes) {
    final int len=bytes.length;
    final char[] chars=new char[len<<1];
    int hexIndex;
    int idx=0;
    int ofs=0;
    while (ofs<len) {
      hexIndex=(bytes[ofs++] & 0xFF)<<1;
      chars[idx++]=BYTE2HEX[hexIndex++];
      chars[idx++]=BYTE2HEX[hexIndex];
    }
    return new String(chars);
  }

@maybewecouldstealavan提出的解决方案的一个小变种，它让你在输出十六进制字符串中可视化地捆绑N个字节:

 final static char[] HEX_ARRAY = "0123456789ABCDEF".toCharArray();
 final static char BUNDLE_SEP = ' ';

public static String bytesToHexString(byte[] bytes, int bundleSize /*[bytes]*/]) {
        char[] hexChars = new char[(bytes.length * 2) + (bytes.length / bundleSize)];
        for (int j = 0, k = 1; j < bytes.length; j++, k++) {
                int v = bytes[j] & 0xFF;
                int start = (j * 2) + j/bundleSize;

                hexChars[start] = HEX_ARRAY[v >>> 4];
                hexChars[start + 1] = HEX_ARRAY[v & 0x0F];

                if ((k % bundleSize) == 0) {
                        hexChars[start + 2] = BUNDLE_SEP;
                }   
        }   
        return new String(hexChars).trim();    
}

那就是:

bytesToHexString("..DOOM..".toCharArray().getBytes(), 2);
2E2E 444F 4F4D 2E2E

bytesToHexString("..DOOM..".toCharArray().getBytes(), 4);
2E2E444F 4F4D2E2E

在存储查找表的很小代价下，这个实现非常简单和快速。

 private static final char[] BYTE2HEX=(
    "000102030405060708090A0B0C0D0E0F"+
    "101112131415161718191A1B1C1D1E1F"+
    "202122232425262728292A2B2C2D2E2F"+
    "303132333435363738393A3B3C3D3E3F"+
    "404142434445464748494A4B4C4D4E4F"+
    "505152535455565758595A5B5C5D5E5F"+
    "606162636465666768696A6B6C6D6E6F"+
    "707172737475767778797A7B7C7D7E7F"+
    "808182838485868788898A8B8C8D8E8F"+
    "909192939495969798999A9B9C9D9E9F"+
    "A0A1A2A3A4A5A6A7A8A9AAABACADAEAF"+
    "B0B1B2B3B4B5B6B7B8B9BABBBCBDBEBF"+
    "C0C1C2C3C4C5C6C7C8C9CACBCCCDCECF"+
    "D0D1D2D3D4D5D6D7D8D9DADBDCDDDEDF"+
    "E0E1E2E3E4E5E6E7E8E9EAEBECEDEEEF"+
    "F0F1F2F3F4F5F6F7F8F9FAFBFCFDFEFF").toCharArray();
   ; 

  public static String getHexString(byte[] bytes) {
    final int len=bytes.length;
    final char[] chars=new char[len<<1];
    int hexIndex;
    int idx=0;
    int ofs=0;
    while (ofs<len) {
      hexIndex=(bytes[ofs++] & 0xFF)<<1;
      chars[idx++]=BYTE2HEX[hexIndex++];
      chars[idx++]=BYTE2HEX[hexIndex];
    }
    return new String(chars);
  }

好的，有很多方法可以做到这一点，但如果你决定使用一个库，我建议在你的项目中看看是否已经在一个库中实现了一些东西，而这个库已经是你项目的一部分，然后再添加一个新的库来做到这一点。例如，如果你还没有

org.apache.commons.codec.binary.Hex

也许你有…

org.apache.xerces.impl.dv.util.HexBin

如果你想让它更具可读性，并将字节彼此分开，你可以在Java 17+中使用以下代码:

byte[] yourByteArray = { -128, 0, 127 };
String hexString = new String(HexFormat.ofDelimiter(" ").formatHex(yourByteArray));
// 80 00 7f

最简单的解决方案，没有外部库，没有数字常量:

public static String byteArrayToHex(byte[] a) {
   StringBuilder sb = new StringBuilder(a.length * 2);
   for(byte b: a)
      sb.append(String.format("%02x", b));
   return sb.toString();
}