实际上，我认为BigInteger是很好的解决方案:

new BigInteger("00A0BF", 16).toByteArray();

编辑:正如海报所指出的，前导零不安全。

一行程序: 进口javax.xml.bind.DatatypeConverter; (字节[]数组) 返回DatatypeConverter.printHexBinary(数组); ｝ public static byte[] toByteArray(String s) { 返回DatatypeConverter.parseHexBinary (s); ｝

对于那些对来自FractalizeR的一行代码背后的实际代码感兴趣的人(我需要它，因为javax.xml.bind不适用于Android(默认情况下))，这来自com.sun.xml.internal.bind. datatypeconverterimpll .java:

public byte[] parseHexBinary(String s) {
    final int len = s.length();

    // "111" is not a valid hex encoding.
    if( len%2 != 0 )
        throw new IllegalArgumentException("hexBinary needs to be even-length: "+s);

    byte[] out = new byte[len/2];

    for( int i=0; i<len; i+=2 ) {
        int h = hexToBin(s.charAt(i  ));
        int l = hexToBin(s.charAt(i+1));
        if( h==-1 || l==-1 )
            throw new IllegalArgumentException("contains illegal character for hexBinary: "+s);

        out[i/2] = (byte)(h*16+l);
    }

    return out;
}

private static int hexToBin( char ch ) {
    if( '0'<=ch && ch<='9' )    return ch-'0';
    if( 'A'<=ch && ch<='F' )    return ch-'A'+10;
    if( 'a'<=ch && ch<='f' )    return ch-'a'+10;
    return -1;
}

private static final char[] hexCode = "0123456789ABCDEF".toCharArray();

public String printHexBinary(byte[] data) {
    StringBuilder r = new StringBuilder(data.length*2);
    for ( byte b : data) {
        r.append(hexCode[(b >> 4) & 0xF]);
        r.append(hexCode[(b & 0xF)]);
    }
    return r.toString();
}

现在可以在guava中使用BaseEncoding来实现这一点。

BaseEncoding.base16().decode(string);

反过来使用

BaseEncoding.base16().encode(bytes);

迟到了，但我已经把DaveL上面的答案合并到一个具有反向操作的类中——以防它有所帮助。

public final class HexString {
    private static final char[] digits = "0123456789ABCDEF".toCharArray();

    private HexString() {}

    public static final String fromBytes(final byte[] bytes) {
        final StringBuilder buf = new StringBuilder();
        for (int i = 0; i < bytes.length; i++) {
            buf.append(HexString.digits[(bytes[i] >> 4) & 0x0f]);
            buf.append(HexString.digits[bytes[i] & 0x0f]);
        }
        return buf.toString();
    }

    public static final byte[] toByteArray(final String hexString) {
        if ((hexString.length() % 2) != 0) {
            throw new IllegalArgumentException("Input string must contain an even number of characters");
        }
        final int len = hexString.length();
        final byte[] data = new byte[len / 2];
        for (int i = 0; i < len; i += 2) {
            data[i / 2] = (byte) ((Character.digit(hexString.charAt(i), 16) << 4)
                    + Character.digit(hexString.charAt(i + 1), 16));
        }
        return data;
    }
}

和JUnit测试类:

public class TestHexString {

    @Test
    public void test() {
        String[] tests = {"0FA1056D73", "", "00", "0123456789ABCDEF", "FFFFFFFF"};

        for (int i = 0; i < tests.length; i++) {
            String in = tests[i];
            byte[] bytes = HexString.toByteArray(in);
            String out = HexString.fromBytes(bytes);
            System.out.println(in); //DEBUG
            System.out.println(out); //DEBUG
            Assert.assertEquals(in, out);

        }

    }

}

这里有另一个版本，它支持奇数长度的字符串，而不诉诸字符串连接。

public static byte[] hexStringToByteArray(String input) {
    int len = input.length();

    if (len == 0) {
        return new byte[] {};
    }

    byte[] data;
    int startIdx;
    if (len % 2 != 0) {
        data = new byte[(len / 2) + 1];
        data[0] = (byte) Character.digit(input.charAt(0), 16);
        startIdx = 1;
    } else {
        data = new byte[len / 2];
        startIdx = 0;
    }

    for (int i = startIdx; i < len; i += 2) {
        data[(i + 1) / 2] = (byte) ((Character.digit(input.charAt(i), 16) << 4)
                + Character.digit(input.charAt(i+1), 16));
    }
    return data;
}

common -codec中的Hex类应该为您做到这一点。

http://commons.apache.org/codec/

import org.apache.commons.codec.binary.Hex;
...
byte[] decoded = Hex.decodeHex("00A0BF");
// 0x00 0xA0 0xBF