在Java中,我有2个字节:
byte b1 = (byte) 0b11111111, b2 = (byte) 0b00000000;
我想混合它们,以便每个第一位来自b1
,而另一位来自b2
(从左到右阅读)。输入的前半部分和后半部分是分开完成的,因此结果是2个字节。结果b3
和b4
如下所示。
byte b3 = (byte) 0b10101010, b4 = 0b10101010;
说明这些位是如何唯一的(使用字母指定唯一位):
byte b1 = (byte) 0bHGFEDCBA, b2 = (byte) 0bPONMLKJI;
结果将是:
byte b3 = (byte) 0bHPGOFNEM, b4 = 0bDLCKBJAI;
或者,图形上,
+---+---+---+---+---+---+---+---+
b1 | H | G | F | E | D | C | B | A |
+---+---+---+---+---+---+---+---+
| | | | | | | |
| | | | | | | +--------------------------------------------+
| | | | | | +----------------------------------------+ |
| | | | | +------------------------------------+ | |
| | | | +--------------------------------+ | | |
| | | +-------------------+ | | | |
| | +---------------+ | | | | |
| +-----------+ | | | | | |
+-------+ | | | | | | |
| | | | | | | |
+---+---+---+---+---+---+---+---+ +---+---+---+---+---+---+---+---+
b3 | H | P | G | O | F | N | E | M | b4 | D | L | C | K | B | J | A | I |
+---+---+---+---+---+---+---+---+ +---+---+---+---+---+---+---+---+
| | | | | | | |
+-----------+ | | | | | | |
| +---------------+ | | | | | |
| | +-------------------+ | | | | |
| | | +-----------------------+ | | | |
| | | | +------------------------------------+ | | |
| | | | | +----------------------------------------+ | |
| | | | | | +--------------------------------------------+ |
| | | | | | | +------------------------------------------------+
| | | | | | | |
+---+---+---+---+---+---+---+---+
b2 | P | O | N | M | L | K | J | I |
+---+---+---+---+---+---+---+---+
实现这一目标的最简单方法是什么?
答案 0 :(得分:1)
正如你所说,如果你的心脏被放在一个单线上,那该怎么样:
public static int interleave(short b1, short b2) {
return((int)(((b2 * 0x0101010101010101L & 0x8040201008040201L) *
0x0102040810204081L >> 49) & 0x5555) |
(int)(((b1 * 0x0101010101010101L & 0x8040201008040201L) *
0x0102040810204081L >> 48) & 0xAAAA));
}
这将返回带有b3&的int。 b4作为低16位,你可以移位和掩盖:
int b3b4 = interleave(b1, b2);
int b3 = b3b4 >> 8;
int b4 = b3b4 & 0b11111111;
提供
答案 1 :(得分:0)
首先,创建一个在一个字节中扩展位的方法,并返回一个int
,其中最后16位设置为原始字节的位:
static int spread(int b) {
int res = 0;
for (int i = 0 ; i != 8 ; i++) {
if ((b & 1<<i) != 0) {
res |= 1<<(2*i);
}
}
return res;
}
使用这个方法,通过对第一个点差的结果进行OR运算产生结果,第二个点差的结果向左移一个:
int res = spread(b1) | (spread(b2) << 1);
由于您的号码很小,您可以为所有256种可能性预先计算spread(x)
。
这produces Morton's table。将其复制到您的班级,并使您的解决方案成为一个单行:
int res = morton[b1] | (morton[b2] << 1);
// This declaration goes at the bottom of your file.
// The numbers are copied from the program output at the link above:
private static final short[] morton = new short[] {
0x0000, 0x0001, 0x0004, 0x0005, 0x0010, 0x0011, 0x0014, 0x0015,
0x0040, 0x0041, 0x0044, 0x0045, 0x0050, 0x0051, 0x0054, 0x0055,
0x0100, 0x0101, 0x0104, 0x0105, 0x0110, 0x0111, 0x0114, 0x0115,
0x0140, 0x0141, 0x0144, 0x0145, 0x0150, 0x0151, 0x0154, 0x0155,
0x0400, 0x0401, 0x0404, 0x0405, 0x0410, 0x0411, 0x0414, 0x0415,
0x0440, 0x0441, 0x0444, 0x0445, 0x0450, 0x0451, 0x0454, 0x0455,
0x0500, 0x0501, 0x0504, 0x0505, 0x0510, 0x0511, 0x0514, 0x0515,
0x0540, 0x0541, 0x0544, 0x0545, 0x0550, 0x0551, 0x0554, 0x0555,
0x1000, 0x1001, 0x1004, 0x1005, 0x1010, 0x1011, 0x1014, 0x1015,
0x1040, 0x1041, 0x1044, 0x1045, 0x1050, 0x1051, 0x1054, 0x1055,
0x1100, 0x1101, 0x1104, 0x1105, 0x1110, 0x1111, 0x1114, 0x1115,
0x1140, 0x1141, 0x1144, 0x1145, 0x1150, 0x1151, 0x1154, 0x1155,
0x1400, 0x1401, 0x1404, 0x1405, 0x1410, 0x1411, 0x1414, 0x1415,
0x1440, 0x1441, 0x1444, 0x1445, 0x1450, 0x1451, 0x1454, 0x1455,
0x1500, 0x1501, 0x1504, 0x1505, 0x1510, 0x1511, 0x1514, 0x1515,
0x1540, 0x1541, 0x1544, 0x1545, 0x1550, 0x1551, 0x1554, 0x1555,
0x4000, 0x4001, 0x4004, 0x4005, 0x4010, 0x4011, 0x4014, 0x4015,
0x4040, 0x4041, 0x4044, 0x4045, 0x4050, 0x4051, 0x4054, 0x4055,
0x4100, 0x4101, 0x4104, 0x4105, 0x4110, 0x4111, 0x4114, 0x4115,
0x4140, 0x4141, 0x4144, 0x4145, 0x4150, 0x4151, 0x4154, 0x4155,
0x4400, 0x4401, 0x4404, 0x4405, 0x4410, 0x4411, 0x4414, 0x4415,
0x4440, 0x4441, 0x4444, 0x4445, 0x4450, 0x4451, 0x4454, 0x4455,
0x4500, 0x4501, 0x4504, 0x4505, 0x4510, 0x4511, 0x4514, 0x4515,
0x4540, 0x4541, 0x4544, 0x4545, 0x4550, 0x4551, 0x4554, 0x4555,
0x5000, 0x5001, 0x5004, 0x5005, 0x5010, 0x5011, 0x5014, 0x5015,
0x5040, 0x5041, 0x5044, 0x5045, 0x5050, 0x5051, 0x5054, 0x5055,
0x5100, 0x5101, 0x5104, 0x5105, 0x5110, 0x5111, 0x5114, 0x5115,
0x5140, 0x5141, 0x5144, 0x5145, 0x5150, 0x5151, 0x5154, 0x5155,
0x5400, 0x5401, 0x5404, 0x5405, 0x5410, 0x5411, 0x5414, 0x5415,
0x5440, 0x5441, 0x5444, 0x5445, 0x5450, 0x5451, 0x5454, 0x5455,
0x5500, 0x5501, 0x5504, 0x5505, 0x5510, 0x5511, 0x5514, 0x5515,
0x5540, 0x5541, 0x5544, 0x5545, 0x5550, 0x5551, 0x5554, 0x5555
};