生成唯一ID，最大长度为3位/字母/符号 [英] Generating unique ids with the max length of 3 digits/letters/simbols
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问题描述
我有75200个单词的列表。我需要为每个单词赋予一个唯一id，每个id的长度可以是3个字母或更少。我可以使用数字，字母甚至符号，但最大长度为3。
I have a list of 75200 words. I need to give a 'unique' id to each word, and the length of each id could be 3 letters or less. I can use numbers, letters or even symbols but the max length is 3.
以下是我的代码。
import java.io.*;
import java.util.*;
public class HashCreator {
private Map completedWordMap;
private String [] simpleLetters = {"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"};
private String[] symbols = {"!","@","#","$","%","^","&","*","~","?"};
private String indexNumber;
String currentlyUsingLetter, currentlyUsingSymbol;
private int currentlyActiveSimpleLetter = 0, currentlyActiveSymbol = 0, currentlyActiveSimpleLetter2 = 0, currentlyActiveSymbol2 = 0;
private boolean secondaryNumberIsHundred = false;
public HashCreator()
{
completedWordMap = createWordNumberingMap();
}
private Map createWordNumberingMap()
{
int number = 0;
int secondaryNumber = 0;
int thirdinoryNumber = 0;
Map wordMap = new HashMap();
BufferedReader br = null;
String str = "";
boolean reset = false;
//First Read The File
File readingFile = new File("WordList/NewWordsList.txt");
try
{
br = new BufferedReader(new FileReader(readingFile));
while((str=br.readLine())!=null)
{
if(number<1000) //Asign numbers from 0 t0 999
{
indexNumber = String.valueOf(number);
wordMap.put(indexNumber, str);
number++;
System.out.println(indexNumber);
}
else // It is 1000 now. Length exceeds so find another way.
{
if(indexNumber.length()<4)
{
if(currentlyActiveSimpleLetter<simpleLetters.length) //Start using simple letter array
{
if(secondaryNumber<100) //Start combining numbers with letters. Results will look like 'a0', a1', 'a2'......'x98',x99'
{
indexNumber = simpleLetters[currentlyActiveSimpleLetter]+secondaryNumber;
wordMap.put(indexNumber, str);
secondaryNumber++;
System.out.println(indexNumber);
}
else
{
//If the number is 100, that means the last result is something like 'a99','b99'...'x99'
//Time to use a new letter and set the counter back to 0 and select the next letter
secondaryNumber = 0;
currentlyActiveSimpleLetter++;
}
}
else
{
if(currentlyActiveSymbol<symbols.length) //We have used the entire alphabet. Start using sybmols now.
{
if(currentlyActiveSymbol==0) //If this is the first time we are reaching this step, reset the counter to 0
{
secondaryNumber = 0;
}
if(secondaryNumber<100)
{
indexNumber = symbols[currentlyActiveSymbol]+secondaryNumber;
wordMap.put(indexNumber, str);
secondaryNumber++;
System.out.println(indexNumber);
}
else
{
//If the number is 100, that means the last result is something like '!99','@99'...'*99'
//Time to use a new letter and set the counter back to 0 and select the next symbol
secondaryNumber = 0;
currentlyActiveSymbol++;
}
}
else
{
//We have used entire list of numbers (0-999), entire list of letters (a0-z99) and entire set of symbols (!0 - ?99)
//Now we need to combine all 3 together.
if(thirdinoryNumber<10)//We are starting with a new 'Number' counter
{
//We again start with replacing numbers. Here the first few and last few results will look like a!0'.....'a!9'
indexNumber = simpleLetters[currentlyActiveSimpleLetter2]+symbols[currentlyActiveSymbol]+thirdinoryNumber;
wordMap.put(indexNumber, str);
thirdinoryNumber++;
System.out.println(indexNumber);
thirdinoryNumber++;
}
else
{
//We have used number from 0-9. Time to start replacing letters
if(currentlyActiveSimpleLetter2<simpleLetters.length)
{
if(currentlyActiveSimpleLetter2==0) //If this is the 'first' time we reach this point, reset the number counter.
{
thirdinoryNumber = 0;
}
if(thirdinoryNumber<10)
{
indexNumber = simpleLetters[currentlyActiveSimpleLetter2]+symbols[currentlyActiveSymbol]+thirdinoryNumber;
wordMap.put(indexNumber, str);
thirdinoryNumber++;
System.out.println(indexNumber);
}
else
{
thirdinoryNumber = 0;
currentlyActiveSimpleLetter2++; //If we are at the peek of usable numbers (0-9) reset simpleletter array position to
// 0 and numbercounter to 0
}
}
else
{
//We have used number from 0-9. Time to start replacing symbols
if(currentlyActiveSymbol2<symbols.length)
{
if(currentlyActiveSymbol2==0) //If this is the 'first' time we reach this point, reset the number counter.
{
thirdinoryNumber = 0;
}
if(thirdinoryNumber<10)
{
indexNumber = simpleLetters[currentlyActiveSimpleLetter2]+symbols[currentlyActiveSymbol]+thirdinoryNumber;
wordMap.put(indexNumber, str);
thirdinoryNumber++;
System.out.println(indexNumber);
}
else
{
thirdinoryNumber = 0;
currentlyActiveSymbol2++; //If we are at the peek of usable numbers (0-9) reset symbol array position to
// 0 and numbercounter to 0
}
}
}
}
}
}
}
else
{
System.out.println("Error in Somewhere. Length Exceeded");
}
}
}
br.close();
System.out.println("Completed");
System.out.println(wordMap.get(0));
}
catch(Exception e)
{
e.printStackTrace();
}
finally
{
try
{
br.close();
}
catch(Exception e)
{
e.printStackTrace();
}
}
return wordMap;
}
}
不幸的是，这不起作用。它打印结果，结果为' z99 '后，它是一堆！0 。下面是它的一小部分：
Unfortunately this doesn't work. It prints the results, and it is bunch of !0 after the result 'z99'. Below is a small piece of it:
!0
!0
!0
!0
...
Completed
null
除了在 k99 之后，它已经生成了10-199的ID，然后正确地从 m0 开始。您可以从此处找到结果文件。
Apart from that, after k99, it has generated ids from 10-199 then started back with m0 properly. You can find the result file from here.
如您所见， wordMap.get（0）也生成 null 。这有什么不对？如果有任何其他简单的方法来生成最多3位数/字母/符号长度的75000个唯一ID，我非常乐意随之移动。
As you can see, wordMap.get(0) also generated null. What is wrong here? If there is any other simple method for generating 75000 unique ids with maximum 3 digits/letters/symbols length, I am more than happy to move with it.
推荐答案
您可以创建一个基本上将十进制数转换为您选择的基数的方法。这里我有46个符号，例如，它给出97336个唯一序列：
You could create a method that basically converts a decimal number to a base of your choice. Here I have 46 symbols for example, which gives 97336 unique sequences:
private static final String[] symbols = { "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 getSequence(final int i) {
return symbols[i / (symbols.length * symbols.length)] + symbols[(i / symbols.length) % symbols.length]
+ symbols[i % symbols.length];
}
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