带上下限的红包算法

群红包总金额m，分n份，每份下限x，上限y，生成一种的随机的可行分配方案

最简单的版本，随机

思路就是先分配最小金额，然后把剩下的金额不断随机加上去，直到为0。这种不太适合流式处理。

public static List<Integer> generateRedPacket(int m, int n, int x, int y) {
    int minTotal = n * x;
    int maxTotal = n * y;
    if (m > maxTotal || m < minTotal) {
        return null;
    }

    int remaining = m - minTotal;
    Random r = new Random();
    List<Integer> redPacket = new ArrayList<>(n);
    for (int i = 0; i < n; i++) {
        redPacket.add(x);
    }

    int idx = 0;
    while (remaining > 0) {
        int maxAddition = Math.min(y - redPacket.get(idx), remaining);
        int addition = r.nextInt(maxAddition + 1);
        redPacket.set(idx, addition + redPacket.get(idx));
        idx++;
        remaining -= addition;
    }

    Collections.shuffle(redPacket, r);
    return redPacket;
}

二倍均值法

能保证生成10个红包

public static List<Integer> generateRedPacket2(int m, int n, int x, int y) {
        int minTotal = n * x;
        int maxTotal = n * y;
        if (m > maxTotal || m < minTotal) {
            return null;
        }

        int remaining = m;
        Random r = new Random();
        List<Integer> redPacket = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            int min = Math.max(x, remaining - (n - i - 1) * y);
            int max = Math.min(y, remaining - (n - i - 1) * x);

            int amount = min + r.nextInt(max - min + 1);
            redPacket.add(amount);
            remaining -= amount;
        }

        Collections.shuffle(redPacket, r);
        return redPacket;
    }