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);
  }

我们不需要使用任何外部库，也不需要基于循环和常量编写代码。 这样就够了:

byte[] theValue = .....
String hexaString = new BigInteger(1, theValue).toString(16);

只是迭代所有字节，转换为十六进制字符串使用整数。toString(不幸的是，没有字节。toString(带基数形参)并将它们全部附加到StringBuilder中。

byte[] arr;//set it to your value
StringBuilder sb=new StringBuilder(arr.length*2);//1 byte...2 hex digits
for(int i=0;i<arr.length;i++){
  sb.append(Integer.toString(arr[i],16));
}
String hexValue=sb.toString();

这类似于指针Null的答案，但它使用整数。toString而不是String。格式提高性能。

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);
  }

我发现了三种不同的方法: http://www.rgagnon.com/javadetails/java-0596.html

他还提到，我认为最优雅的一个是:

static final String HEXES = "0123456789ABCDEF";
public static String getHex( byte [] raw ) {
    if ( raw == null ) {
        return null;
    }
    final StringBuilder hex = new StringBuilder( 2 * raw.length );
    for ( final byte b : raw ) {
        hex.append(HEXES.charAt((b & 0xF0) >> 4))
            .append(HEXES.charAt((b & 0x0F)));
    }
    return hex.toString();
}

  public static byte[] hexStringToByteArray(String s) {
    int len = s.length();
    byte[] data = new byte[len / 2];
    for (int i = 0; i < len; i += 2) {
      data[i / 2] = (byte) ((Character.digit(s.charAt(i), 16) << 4)
        + Character.digit(s.charAt(i+1), 16));
    }
  return data;
  }