java进制转换

对于10进制数转换为N（2-36）进制一般都是选择取余除的算法进行转换 ，下面给出两种方案

一种是递归，一种是迭代。通过效率评价两者性能

其中迭代的方案直接取自java源代码。

/*
*Class NotationConvert.java
*Create Date: 2009-11-12
*Author:a276202460
*/
package com.rich.notation;

public class NotationConvert {
private static final String[] letters = { "0", "1", "2", "3", "4", "5",
"6", "7", "8", "9", "a", "b", "c", "d", "e", "f", "g", "h", "i",
"g", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v",
"w", "x", "y", "z" };

static final char[] digits = { '0', '1', '2', '3', '4', '5', '6', '7', '8',
'9', 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l',
'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y',
'z' };

public static String TentoN(int value, int number) {
if (number <= 1 || number > letters.length) {
throw new RuntimeException("Faild");
}
if (value < 0) {
return "-" + TentoN(0 - value, number);
}
if (value < number) {
return letters[value];
} else {
return (TentoN(value / number, number) + letters[value % number]);
}
}

public static String TentoN1(int value, int number) {
if (number <= 1 || number > letters.length) {
throw new RuntimeException("Faild");
}
char[] rs = new char[33];
boolean flag = value < 0;
int startindex = 32;
if (!flag)
value = 0 - value;
for (; value <= -number; value /= number)
rs[startindex--] = digits[-(value % number)];
rs[startindex] = digits[-value];
if (flag) {
rs[--startindex] = '-';
}
return new String(rs, startindex, 33 - startindex);
}

public static void main(String[] s) {
long starttime = System.currentTimeMillis();
for (int i = 0; i < 100000; i++)
TentoN(-33, 2);
System.out.println((System.currentTimeMillis() - starttime));
starttime = System.currentTimeMillis();
for (int i = 0; i < 100000; i++)
TentoN1(-33, 2);
System.out.println((System.currentTimeMillis() - starttime));

}
}

其中TentoN方法使用了递归的方案进行求解，TentoN1方法使用迭代求解。方案1使用了字符串相加性能上自然会掉一大截，二迭代的方法中经过计算后只需修改char数组的制定索引位置的值，要快的多。

此程序在本机运行的结果为：

188
31

速度上6倍左右。