java - 如何生成4个没有重复数字的随机数

Question

我想生成 4 随机数，它们不能有重复的数字。

例如 4567 ，(它没有像 4557 这样的重复值)。

我希望它们是 随机 。

有没有办法实现这一目标？

Answer 1

我现在沉迷于流。但是流很慢。为了显示我写了以下主要方法的缓慢。 “generateWithLoop”方法涵盖了@WJS 的答案。

    public static void main(String[] args) {

        long nanoStart = System.nanoTime();
        generateWithStreams();
        long nanoEnd = System.nanoTime();
        System.out.println("Elapsed time with Streams : " + (nanoEnd - nanoStart) + " nano seconds");

        nanoStart = System.nanoTime();
        generateWithLoop();
        nanoEnd = System.nanoTime();
        System.out.println("Elapsed time with Loop : " + (nanoEnd - nanoStart) + " nano seconds");
    }

控制台输出:

Elapsed time with Streams : 81367089 nano seconds

Elapsed time with Loop : 75093 nano seconds

随着流:

    public static void generateWithStreams() {

        List<Integer> orderedList = getOrderedList();

        for (int i = 0; i < 4; i++) {
            List<Integer> shuffledList = getShuffledList(orderedList);
            System.out.println(get4DigitNumber(shuffledList));
        }

    }

    public static List<Integer> getOrderedList() {
        return IntStream.range(0, 10).boxed().collect(Collectors.toList());
    }

    public static List<Integer> getShuffledList(List<Integer> list) {
        return list.stream().sorted((o1, o2) -> ThreadLocalRandom.current().nextInt(-1, 2)).collect(Collectors.toList());
    }

    public static Integer get4DigitNumber(List<Integer> shuffledList) {
        final Integer LIMIT = shuffledList.get(0).equals(0) ? 5 : 4;
        return shuffledList.stream().limit(LIMIT).reduce(0, (sum, current) -> sum * 10 + current);
    }

带循环:

    public static void generateWithLoop() {
        Random r = new Random();
        for (int k = 0; k < 4; k++) {
            int val = gen(r);
            System.out.println(val);
        }

    }

    static int[] base = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };

    public static int gen(Random r) {
        int val = 0;
        int s = 10;
        for (int i = 0; i < 4; i++) {
            int n = r.nextInt(s);
            val = val * 10 + base[n];
            int save = base[n];
            base[n] = base[--s];
            base[s] = save;
        }
        return val < 1000 ? val * 10 + base[r.nextInt(s)] : val;
    }