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wolf_awp
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由UUID和短域名想到的

ITeyeCC++C#F#
阅读更多
帖子地址：http://www.iteye.com/topic/1017979
UUID是把128个二进制数，转换成32个16进制数的，每4个二进制数转换成一个16进制数。

如果是64（2的6次方）进制的话，应该是6个二进制数转换一个64进制数。

我们可以在UUID前面补加一个16进行数，让它成为33位的16进制数，共是132位二进制数。

这样就可以用22个64进制数表示132（22*6）位的二进制数。

结论是，可以把36位的UUID，去掉“-”变成32位的16进制数。
在这个数前面补一个16进制数，比如“0”，就变成了33位（132个二进制数，33*4）。
再把这个33位的16进制数，转换成22位的64进制数。

最终是把36位的UUID，变成一个22位的数表示，短了14位。

可略过以下代码，直接看5楼回复。
5楼代码比这优雅的多！ 

public class UUIDDemo {

	public static void main(String[] args) {
		String uuid = "1838efee-893d-3778-b480-43d6457767ea";
		System.out.println("原始uuid：" + uuid);

		uuid = uuid.replaceAll("-", "").toUpperCase();
		System.out.println("去掉'-'变大写:" + uuid);

		uuid = "0" + uuid;
		System.out.println("前补16进制'0':" + uuid);
		
		uuid = hexToBinary(uuid);
		System.out.println("转换为132位2进制数:" + uuid);
		
		uuid = binaryTo64(uuid);
		System.out.println("转换为64进制数:" + uuid);
	}

	public static String hexToBinary(String hex) {
		StringBuffer sb = new StringBuffer();
		if (hex == null || "".equals(hex))
			return "";

		for (int i = 0; i < hex.length(); i++) {
			sb.append(hexToBinary(hex.charAt(i)));
		}

		return sb.toString();
	}

	public static String hexToBinary(char hex) {
		String binary = null;

		switch (hex) {
		case '0':
			binary = "0000";
			break;
		case '1':
			binary = "0001";
			break;
		case '2':
			binary = "0010";
			break;
		case '3':
			binary = "0011";
			break;
		case '4':
			binary = "0100";
			break;
		case '5':
			binary = "0101";
			break;
		case '6':
			binary = "0110";
			break;
		case '7':
			binary = "0111";
			break;
		case '8':
			binary = "1000";
			break;
		case '9':
			binary = "1001";
			break;
		case 'A':
			binary = "1010";
			break;
		case 'B':
			binary = "1011";
			break;
		case 'C':
			binary = "1100";
			break;
		case 'D':
			binary = "1101";
			break;
		case 'E':
			binary = "1110";
			break;
		case 'F':
			binary = "1111";
			break;
		default:
			binary = "";
		}
		return binary;
	}

	public static String binaryTo64(String binary) {
		StringBuffer sb = new StringBuffer();
		if (binary == null || "".equals(binary))
			return "";
		if ((binary.length() % 6) != 0)
			return "";

		for (int i = 0; i < binary.length(); i += 6) {
			sb.append(binaryTo64a(binary.substring(i, i + 6)));
		}

		return sb.toString();
	}

	public static String binaryTo64a(String binary) {
		String result = null;
		if ("000000".equals(binary)) {
			result = "0";
		} else if ("000001".equals(binary)) {
			result = "1";
		} else if ("000010".equals(binary)) {
			result = "2";
		} else if ("000011".equals(binary)) {
			result = "3";
		} else if ("000100".equals(binary)) {
			result = "4";
		} else if ("000101".equals(binary)) {
			result = "5";
		} else if ("000110".equals(binary)) {
			result = "6";
		} else if ("000111".equals(binary)) {
			result = "7";
		} else if ("001000".equals(binary)) {
			result = "8";
		} else if ("001001".equals(binary)) {
			result = "9";
		} else if ("001010".equals(binary)) {
			result = "a";
		} else if ("001011".equals(binary)) {
			result = "b";
		} else if ("001100".equals(binary)) {
			result = "c";
		} else if ("001101".equals(binary)) {
			result = "d";
		} else if ("001110".equals(binary)) {
			result = "e";
		} else if ("001111".equals(binary)) {
			result = "f";
		} else if ("010000".equals(binary)) {
			result = "g";
		} else if ("010001".equals(binary)) {
			result = "h";
		} else if ("010010".equals(binary)) {
			result = "i";
		} else if ("010011".equals(binary)) {
			result = "j";
		} else if ("010100".equals(binary)) {
			result = "k";
		} else if ("010101".equals(binary)) {
			result = "l";
		} else if ("010110".equals(binary)) {
			result = "m";
		} else if ("010111".equals(binary)) {
			result = "n";
		} else if ("011000".equals(binary)) {
			result = "o";
		} else if ("011001".equals(binary)) {
			result = "p";
		} else if ("011010".equals(binary)) {
			result = "q";
		} else if ("011011".equals(binary)) {
			result = "r";
		} else if ("011100".equals(binary)) {
			result = "s";
		} else if ("011101".equals(binary)) {
			result = "t";
		} else if ("011110".equals(binary)) {
			result = "u";
		} else if ("011111".equals(binary)) {
			result = "v";
		} else if ("100000".equals(binary)) {
			result = "w";
		} else if ("100001".equals(binary)) {
			result = "x";
		} else if ("100010".equals(binary)) {
			result = "y";
		} else if ("100011".equals(binary)) {
			result = "z";
		} else if ("100100".equals(binary)) {
			result = "A";
		} else if ("100101".equals(binary)) {
			result = "B";
		} else if ("100110".equals(binary)) {
			result = "C";
		} else if ("100111".equals(binary)) {
			result = "D";
		} else if ("101000".equals(binary)) {
			result = "E";
		} else if ("101001".equals(binary)) {
			result = "F";
		} else if ("101010".equals(binary)) {
			result = "G";
		} else if ("101011".equals(binary)) {
			result = "H";
		} else if ("101100".equals(binary)) {
			result = "I";
		} else if ("101101".equals(binary)) {
			result = "J";
		} else if ("101110".equals(binary)) {
			result = "K";
		} else if ("101111".equals(binary)) {
			result = "L";
		} else if ("110000".equals(binary)) {
			result = "M";
		} else if ("110001".equals(binary)) {
			result = "N";
		} else if ("110010".equals(binary)) {
			result = "O";
		} else if ("110011".equals(binary)) {
			result = "P";
		} else if ("110100".equals(binary)) {
			result = "Q";
		} else if ("110101".equals(binary)) {
			result = "R";
		} else if ("110110".equals(binary)) {
			result = "S";
		} else if ("110111".equals(binary)) {
			result = "T";
		} else if ("111000".equals(binary)) {
			result = "U";
		} else if ("111001".equals(binary)) {
			result = "V";
		} else if ("111010".equals(binary)) {
			result = "W";
		} else if ("111011".equals(binary)) {
			result = "X";
		} else if ("111100".equals(binary)) {
			result = "Y";
		} else if ("111101".equals(binary)) {
			result = "Z";
		} else if ("111110".equals(binary)) {
			result = "_";
		} else if ("111111".equals(binary)) {
			result = "-";
		} else {
			result = "";
		}

		return result;
	}
}


输出结果: 
原始uuid：1838efee-893d-3778-b480-43d6457767ea
去掉'-'变大写:1838EFEE893D3778B48043D6457767EA
前补16进制'0':01838EFEE893D3778B48043D6457767EA
转换为132位2进制数:
000000011000001110001110111111101110100010010011110100110111011110001011010010000000010000111101011001000101011101110110011111101010
转换为64进制数:0oee-KyjQTubi0gZp5tSvG
分享到：
| java约瑟夫环
评论
10 楼 big1980 2014-07-11  
  
9 楼 524386747 2013-10-15  
这域名也不短哈
8 楼 wolf_awp 2011-04-27  
aochant 写道
要实现这个目的，枚举是不可避免的

5楼大牛的代码已经说明，要实现个这目的，枚举是可以避免的
7 楼 aochant 2011-04-27  
要实现这个目的，枚举是不可避免的
6 楼 wolf_awp 2011-04-27  
william_ai 写道
public class Test {
	public static final char[] charMap;
	static {
		charMap = new char[64];
		for (int i = 0; i < 10; i++) {
			charMap[i] = (char) ('0' + i);
		}
		for (int i = 10; i < 36; i++) {
			charMap[i] = (char) ('a' + i - 10);
		}
		for (int i = 36; i < 62; i++) {
			charMap[i] = (char) ('A' + i - 36);
		}
		charMap[62] = '_';
		charMap[63] = '-';
	}

	public static String hexTo64(String hex) {
		StringBuffer r = new StringBuffer();
		int index = 0;
		int[] buff = new int[3];
		int l = hex.length();
		for (int i = 0; i < l; i++) {
			index = i % 3;
			buff[index] = Integer.parseInt(""+hex.charAt(i),16);
			if (index == 2) {
				r.append(charMap[buff[0] << 2 | buff[1] >>> 2]);
				r.append(charMap[(buff[1] &3) << 4 | buff[2]]);
			}
		}
		return r.toString();
	}

	public static void process() {
		String uuid = "1838efee-893d-3778-b480-43d6457767ea";
		uuid = uuid.replaceAll("-", "").toUpperCase();
		uuid = "0" + uuid;
		uuid = hexTo64(uuid);
	}
	
	public static void test() {
		for (int i = 0; i < 100000; i++) {
			process();
		}
		long t1 = System.nanoTime();
		for (int i = 0; i < 100000; i++) {
			process();
		}
		long t2 = System.nanoTime();
		System.out.println(t2 - t1);
	}

	public static void main(String[] args) {
		test();
	}
}

我也想应该有这种方法，只是技术有限，没这样写。
5 楼 william_ai 2011-04-27   
public class Test {
	public static final char[] charMap;
	static {
		charMap = new char[64];
		for (int i = 0; i < 10; i++) {
			charMap[i] = (char) ('0' + i);
		}
		for (int i = 10; i < 36; i++) {
			charMap[i] = (char) ('a' + i - 10);
		}
		for (int i = 36; i < 62; i++) {
			charMap[i] = (char) ('A' + i - 36);
		}
		charMap[62] = '_';
		charMap[63] = '-';
	}

	public static String hexTo64(String hex) {
		StringBuffer r = new StringBuffer();
		int index = 0;
		int[] buff = new int[3];
		int l = hex.length();
		for (int i = 0; i < l; i++) {
			index = i % 3;
			buff[index] = Integer.parseInt(""+hex.charAt(i),16);
			if (index == 2) {
				r.append(charMap[buff[0] << 2 | buff[1] >>> 2]);
				r.append(charMap[(buff[1] &3) << 4 | buff[2]]);
			}
		}
		return r.toString();
	}

	public static void process() {
		String uuid = "1838efee-893d-3778-b480-43d6457767ea";
		uuid = uuid.replaceAll("-", "").toUpperCase();
		uuid = "0" + uuid;
		uuid = hexTo64(uuid);
	}
	
	public static void test() {
		for (int i = 0; i < 100000; i++) {
			process();
		}
		long t1 = System.nanoTime();
		for (int i = 0; i < 100000; i++) {
			process();
		}
		long t2 = System.nanoTime();
		System.out.println(t2 - t1);
	}

	public static void main(String[] args) {
		test();
	}
}
4 楼 crskyp 2011-04-27  
sdh5724 写道
这代码。。。我哭了， 枚举法。。

同意枚举
3 楼 qq376727939 2011-04-27  
也太繁琐了吧！
2 楼 287854442 2011-04-27  
sdh5724 写道
这代码。。。我哭了， 枚举法。。

枚举法看似笨拙 其实效率是很高的
1 楼 sdh5724 2011-04-27  
这代码。。。我哭了， 枚举法。。
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