在某些应用场景下,需要对数字进行加密,skip32无疑是一种很实用的算法,网上有python的源码(https://bitbucket.org/anuraguniyal/skip32.py/src/403577f22bc2300b30b746526f52b6cf3c8cad51/skip32.py?at=master),可惜没有java的版本。下面送上自己写的java版本:
import java.nio.ByteBuffer; import java.nio.ByteOrder; public class Skip32 { public static void main(String[] args) { long number = 4294967295L; String key = "djwiji6789"; String enumber = skip32encrypt(number, key); Long dnumber = skip32decrypt(enumber, key); System.out.println(enumber); System.out.println(dnumber); } public static String skip32encrypt(long number, String key) { return Long.toString(skip32(number, key, true), 36); } public static long skip32decrypt(String base36text, String key) { return skip32(Long.parseLong(base36text, 36), key, false); } private static long skip32(long number, String key, boolean encrypt) { ByteBuffer buffer = ByteBuffer.allocate(8); buffer.order(ByteOrder.LITTLE_ENDIAN); buffer.putLong(number); byte[] b = buffer.array(); int[] a = new int[b.length]; for (int i = 0; i < b.length; i++) { a[i] = b[i] & 0xFF; } int buf[] = new int[4]; for (int i = 0; i < 4; i++) { buf[i] = a[i]; } int[] k = new int[10]; for (int i = 0; i < k.length; i++) { k[i] = key.codePointAt(i); } _skip32(k, buf, encrypt); int index = 0; int firstByte = (0x000000FF & buf[index]); int secondByte = (0x000000FF & buf[index + 1]); int thirdByte = (0x000000FF & buf[index + 2]); int fourthByte = (0x000000FF & buf[index + 3]); long anUnsignedInt = ((long) (fourthByte << 24 | thirdByte << 16 | secondByte << 8 | firstByte)) & 0xFFFFFFFFL; return anUnsignedInt; } private static void _skip32(int[] key, int[] buf, boolean encrypt) { int k; int i; int kstep; int wl, wr; if (encrypt) { kstep = 1; k = 0; } else { kstep = -1; k = 23; } wl = ((buf[0] << 8) + buf[1]); wr = ((buf[2] << 8) + buf[3]); for (i = 0; i < 24 / 2; ++i) { wr ^= g(key, k, wl) ^ k; k += kstep; wl ^= g(key, k, wr) ^ k; k += kstep; } buf[0] = (wr >> 8); buf[1] = (wr & 0xFF); buf[2] = (wl >> 8); buf[3] = (wl & 0xFF); } private static int g(int[] key, int k, int w) { int g1, g2, g3, g4, g5, g6; g1 = ((w >> 8) & 0xff); g2 = (w & 0xff); g3 = (F[g2 ^ key[(4 * k) % 10]] ^ g1); g4 = (F[g3 ^ key[(4 * k + 1) % 10]] ^ g2); g5 = (F[g4 ^ key[(4 * k + 2) % 10]] ^ g3); g6 = (F[g5 ^ key[(4 * k + 3) % 10]] ^ g4); return ((g5 << 8) + g6); } private static final int[] F = new int[] { 0xa3, 0xd7, 0x09, 0x83, 0xf8, 0x48, 0xf6, 0xf4, 0xb3, 0x21, 0x15, 0x78, 0x99, 0xb1, 0xaf, 0xf9, 0xe7, 0x2d, 0x4d, 0x8a, 0xce, 0x4c, 0xca, 0x2e, 0x52, 0x95, 0xd9, 0x1e, 0x4e, 0x38, 0x44, 0x28, 0x0a, 0xdf, 0x02, 0xa0, 0x17, 0xf1, 0x60, 0x68, 0x12, 0xb7, 0x7a, 0xc3, 0xe9, 0xfa, 0x3d, 0x53, 0x96, 0x84, 0x6b, 0xba, 0xf2, 0x63, 0x9a, 0x19, 0x7c, 0xae, 0xe5, 0xf5, 0xf7, 0x16, 0x6a, 0xa2, 0x39, 0xb6, 0x7b, 0x0f, 0xc1, 0x93, 0x81, 0x1b, 0xee, 0xb4, 0x1a, 0xea, 0xd0, 0x91, 0x2f, 0xb8, 0x55, 0xb9, 0xda, 0x85, 0x3f, 0x41, 0xbf, 0xe0, 0x5a, 0x58, 0x80, 0x5f, 0x66, 0x0b, 0xd8, 0x90, 0x35, 0xd5, 0xc0, 0xa7, 0x33, 0x06, 0x65, 0x69, 0x45, 0x00, 0x94, 0x56, 0x6d, 0x98, 0x9b, 0x76, 0x97, 0xfc, 0xb2, 0xc2, 0xb0, 0xfe, 0xdb, 0x20, 0xe1, 0xeb, 0xd6, 0xe4, 0xdd, 0x47, 0x4a, 0x1d, 0x42, 0xed, 0x9e, 0x6e, 0x49, 0x3c, 0xcd, 0x43, 0x27, 0xd2, 0x07, 0xd4, 0xde, 0xc7, 0x67, 0x18, 0x89, 0xcb, 0x30, 0x1f, 0x8d, 0xc6, 0x8f, 0xaa, 0xc8, 0x74, 0xdc, 0xc9, 0x5d, 0x5c, 0x31, 0xa4, 0x70, 0x88, 0x61, 0x2c, 0x9f, 0x0d, 0x2b, 0x87, 0x50, 0x82, 0x54, 0x64, 0x26, 0x7d, 0x03, 0x40, 0x34, 0x4b, 0x1c, 0x73, 0xd1, 0xc4, 0xfd, 0x3b, 0xcc, 0xfb, 0x7f, 0xab, 0xe6, 0x3e, 0x5b, 0xa5, 0xad, 0x04, 0x23, 0x9c, 0x14, 0x51, 0x22, 0xf0, 0x29, 0x79, 0x71, 0x7e, 0xff, 0x8c, 0x0e, 0xe2, 0x0c, 0xef, 0xbc, 0x72, 0x75, 0x6f, 0x37, 0xa1, 0xec, 0xd3, 0x8e, 0x62, 0x8b, 0x86, 0x10, 0xe8, 0x08, 0x77, 0x11, 0xbe, 0x92, 0x4f, 0x24, 0xc5, 0x32, 0x36, 0x9d, 0xcf, 0xf3, 0xa6, 0xbb, 0xac, 0x5e, 0x6c, 0xa9, 0x13, 0x57, 0x25, 0xb5, 0xe3, 0xbd, 0xa8, 0x3a, 0x01, 0x05, 0x59, 0x2a, 0x46 }; }
运行结果:
1ev336c 4294967295
使用注意:以上算法只对无符号32位数字有效。超过32位,可以参考https://github.com/dstar4138/jskipjack/blob/master/src/cipher/SkipJack.java