java - 支持大整数的单词列表生成器

Question

我需要帮助来完成一个程序，该程序将从所选字符和长度生成单词列表(它需要支持大长度)。

首先，您需要通过添加所需的长度(字长)并制作指定字符(字母)的字符串来解决此问题。

所以完整的字数是:

long MAX_WORDS = (long) Math.pow(alphabet.length(), wordlength);

实际上，我做到了并且它有效(例如 2 或 66 个字符的短单词)。

import java.math.BigInteger;
public class wordlistgenenreg {

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

private static void generate(){
int wordlength =2;
String alphabet = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ.-_~";
final long MAX_WORDS = (long) Math.pow(alphabet.length(), wordlength);
final int RADIX = alphabet.length();

for (long i = 0; i < MAX_WORDS; i++) {
    int[] indices = convertToRadix(RADIX, i, wordlength);
    char[] word = new char[wordlength];
    for (int k = 0; k < wordlength; k++) {word[k] = alphabet.charAt(indices[k]);}
    String fullword=new String(word);
    System.out.println(fullword);
}

System.out.println("completed!");
}

private static int[] convertToRadix(int radix, long number, int wordlength) {
int[] indices = new int[wordlength];
for (int i = wordlength - 1; i >= 0; i--) {
    if (number > 0) {
        int rest = (int) (number % radix);
        number /= radix;
        indices[i] = rest;
    } else {
        indices[i] = 0;
    }

}
return indices;
}
}

但是当我想从 66 个字符生成一个非常大的 64 个字符的字符串时，就会出现问题。因为:

MAX_WORDS = 66^64 = 282365657377235405270307754780751252031361330095689004197961218014051357270480550051149871489969454245263206971867136

因此我尝试更改它以使其与 BigInteger 一起使用。但我们的结果是，我总是获得字符串:

"0000000000000000000000000000000000000000000000000000000000000000"

所以有一个问题我没有弄清楚。这是我改变它的工作:

import java.math.BigInteger;

public class wordlistgen {

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

private static void generate() {
int wordlength = 64;
String alphabet = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ.-_~";
BigInteger max_words=new BigInteger("282365657377235405270307754780751252031361330095689004197961218014051357270480550051149871489969454245263206971867136");
final int RADIX = alphabet.length(); 
BigInteger plus=BigInteger.valueOf(1);

for (BigInteger i = new BigInteger("0"); i.compareTo(max_words) <0; i.add(plus)) {
    int[] indices = convertToRadix(RADIX, i, wordlength);
    char[] word = new char[wordlength];
    for (int k = 0; k < wordlength; k++) {word[k] = alphabet.charAt(indices[k]);}
    String fullword=new String(word);
    System.out.println(fullword);        
}
}

private static int[] convertToRadix(int radix, BigInteger i2, int wordlength) {
BigInteger zero=BigInteger.valueOf(0);
BigInteger big_radix=BigInteger.valueOf(radix);
int[] indices = new int[wordlength];
for (int i = wordlength - 1; i >= 0; i--) {
    if (i2.compareTo(zero)==0) {

        BigInteger rest =i2.remainder(big_radix);
        BigInteger ab=i2.divide(big_radix);
        i2=ab;
        indices[i] = rest.intValue();
    } else {
        indices[i] = 0;
    }
}
return indices;
}
}

Answer 1

这是原始版本中的 if:

if (number > 0) {
    int rest = (int) (number % radix);
    number /= radix;
    indices[i] = rest;
} else {
    indices[i] = 0;
}

以及 BigInteger 版本中相同的 if:

if (i2.compareTo(zero)==0) {

    BigInteger rest =i2.remainder(big_radix);
    BigInteger ab=i2.divide(big_radix);
    i2=ab;
    indices[i] = rest.intValue();
} else {
    indices[i] = 0;
}

如您所见，在新的 if 中，您询问是否 number == 0 而不是 number > 0。所以你总是会陷入else。

附注:您正在运行一个从 0 到 max_words 的循环。如果每次迭代只需要一纳秒即可完成，则仍需要 3687886676721203490906725006126388162312177668963067239285600631885632818310441214790267460959878872632642 65 年。足够的时间让宇宙分解成完全熵。我建议重新考虑你的算法。