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81进制,用多进制方式把一个长长的整数变短

时间:2015-07-28 17:48:31      阅读:141      评论:0      收藏:0      [点我收藏+]

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最近在做项目有些资源要用到唯一的标识code,这个code要全局唯一,因此比较长,有25位,long只能处理到19位。另外25位长的一个整数阅读显示都不是很理想,因此开发了一个多进制的转换类。思想接近把一个域名用几个字符代替。下面就是实现的具体java代码,目前支持到81进制。

 1 import java.math.BigInteger;
 2 import java.util.Arrays;
 3 import java.util.Date;
 4 
 5 /**
 6  * 
 7  * @author 程序员老刘@2015-07-28
 8  * @Description 把十进制的数字转换成36、62或者81进制,使表达的长度变短,例如99999转成62进制是Q0t
 9  */
10 public class MultiNumberation {
11 
12     private final static char[] flag = { ‘0‘, ‘1‘, ‘2‘, ‘3‘, ‘4‘, ‘5‘, ‘6‘, ‘7‘, ‘8‘, ‘9‘, ‘A‘,
13             ‘B‘, ‘C‘, ‘D‘, ‘E‘, ‘F‘, ‘G‘, ‘H‘, ‘I‘, ‘J‘, ‘K‘, ‘L‘, ‘M‘, ‘N‘, ‘O‘, ‘P‘, ‘Q‘, ‘R‘,
14             ‘S‘, ‘T‘, ‘U‘, ‘V‘, ‘W‘, ‘X‘, ‘Y‘, ‘Z‘, ‘a‘, ‘b‘, ‘c‘, ‘d‘, ‘e‘, ‘f‘, ‘g‘, ‘h‘, ‘i‘,
15             ‘j‘, ‘k‘, ‘l‘, ‘m‘, ‘n‘, ‘o‘, ‘p‘, ‘q‘, ‘r‘, ‘s‘, ‘t‘, ‘u‘, ‘v‘, ‘w‘, ‘x‘, ‘y‘, ‘z‘,
16             ‘!‘, ‘#‘, ‘%‘, ‘&‘, ‘$‘, ‘*‘, ‘(‘, ‘)‘, ‘[‘, ‘]‘, ‘{‘, ‘}‘, ‘^‘, ‘~‘, ‘?‘, ‘@‘, ‘>‘,
17             ‘<‘, ‘=‘ };
18 
19     public static int getDeep() {
20         return flag.length;
21     }
22 
23     public static int findInFlag(String f) {
24         char c = (char) f.getBytes()[0];
25         for (int i = 0; i < flag.length; i++) {
26             if (flag[i] == c)
27                 return i;
28         }
29         return -1;
30     }
31 
32     public static BigInteger toDecimal(String multi, int deep) {
33         BigInteger result = new BigInteger("0");
34         if (deep < 1 || deep > flag.length || deep == 10)
35             return result;
36 
37         BigInteger d = BigInteger.valueOf(deep);
38         for (int i = 0; i < multi.length(); i++) {
39             int pos = findInFlag(multi.substring(i, i + 1));
40             // Arrays.binarySearch(flag, (char) (multi.substring(i, i +
41             // 1).getBytes()[0]));
42 
43             result = result.add(d.pow(multi.length() - i - 1).multiply(BigInteger.valueOf(pos)));
44         }
45         return result;
46     }
47 
48     /**
49      * 
50      * @Title:
51      * @Description:将十进制整数转为指定进制的数
52      * @param decimal
53      *            --十进制整数
54      * @param deep
55      *            --选择进制,从2~81
56      * @return
57      */
58     public static String toMulti(BigInteger decimal, int deep) {
59         if (deep < 1 || deep > flag.length || deep == 10)
60             return "";
61         // 取余数
62         BigInteger d = BigInteger.valueOf(deep);
63         BigInteger[] bigDivide = decimal.divideAndRemainder(d);
64         int remainder = bigDivide[1].intValue();
65         String result = "" + flag[(int) remainder];
66 
67         // 取商
68         BigInteger quotient = bigDivide[0];
69         // 商数下雨指定的进制数则继续
70         if (quotient.compareTo(d) >= 0) {
71             result = toMulti(quotient, deep) + result;
72         } else {
73             result = "" + flag[(int) quotient.intValue()] + result;
74         }
75 
76         return result;
77     }
78 
79     public static void main(String[] args) {
80         int deep = MultiNumberation.getDeep();
81         // System.out.println("deep=" + deep);
82         long s = new Date().getTime();
83         // MultiNumeration.toMulti(238328L, deep);
84         for (long i = 1000000; i < 1000002; i++) {
85             String code = String.valueOf(i);
86             // String code =
87             // CreateCodeHelper.createCode(SeqObjectName.ServiceGroup_Category);
88             // System.out.println("code=" + code);
89             BigInteger big = new BigInteger(code);
90 
91             String ret = MultiNumberation.toMulti(big, deep);
92             System.out.println(code + ">>" + ret + "===>" + MultiNumberation.toDecimal(ret, deep));
93         }
94         System.out.println("deep=" + deep + ",timeout:" + (new Date().getTime() - s));
95     }
96 }

 

81进制,用多进制方式把一个长长的整数变短

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原文地址:http://www.cnblogs.com/liughost/p/4683214.html

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