java - 可以生成多少个唯一 ID

Question

我正在使用下面的代码创建一个唯一的 Id，它是 8 个字符(包括数字和字母数字字符)。

try {
            List<String> uuidList = new ArrayList<String>();
            int counter = 1;
            File file = new File("D://temp//temp1.txt");
            file.createNewFile();

            Writer writer = new FileWriter(file);
            BufferedWriter wr = new  BufferedWriter(writer);
            while(true) {
                int length = bitsArray.length;
                Random r = new Random();
                StringBuffer uuid = new StringBuffer();
                for(int i= 0; i < 8; i++) {
                    int nextRandomId = r.nextInt(length);
                    uuid.append(bitsArray[nextRandomId]);
                }
                String uuidString = uuid.toString();
                wr.write(uuidString);
                wr.newLine();
                if(counter != 1 && uuidList.contains(uuidString)) {
                    Thread.sleep(1000);
                    System.err.println(counter);
                    break;
                }
                //061e735145fc
                System.err.println(uuidString);
                uuidList.add(uuidString);
                counter++;
            }
        } catch (Exception e) {

        }

我需要知道，如果我使用上面的代码.. 那么我可以生成多少个唯一 ID。给定

static String[] bitsArray = {"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",
                          "0","1","2","3","4","5","6","7","8","9"};

请帮忙..

Answer 1

本质上，您可以生成的字符串总数为 36⁸。

这个定理用Discrete Mathematics(使用Bit String)来解释:

你有 8 个字符的位串，你需要从 36 个字符中选择 1 个来填充:

__  __  __  __  __  __  __  __  
36  36  36  36  36  36  36  36  (characters a -- z, 0-- 9)

因此你有 36*36*36*36*36*36*36*36 = 36⁸= 2,821,109,907,456 总 ID。