Question

我试图从一组数字中得到非重复的随机数。每次我尝试从该数组中获取随机值时，它应该给我非重复随机数。不应重复先前的随机值

int[] integer_array = {0,1,2,3,4,5,6,7};
int random_no = should be random number from above array
int random_no2 = should be random number from above array other than random_no
int random_no3 = should be random number from above array other than random_no
                                                                and random_no2

可以为integer_array.length次生成数组的随机数。

Answer 1

这是我的代码：

public static int[] getIndices(int maxValue, int numberOfIndices) {
    // The result array.
    int[] res = new int[numberOfIndices];
    Random rng = new Random();
    // A set of already used numbers.
    TreeSet<Integer> was = new TreeSet<>();
    for (int i = 0; i < numberOfIndices; i++) {
        // Generate a new number in range [0..maxValue - i].
        // It is a position of a new index in an array of unused values.
        int cur = rng.nextInt(maxValue - i);
        // Compute its position taking into account all values(used and unused)
        // to obtain the real index.
        for (int prev : was)
            if (cur >= prev)
                cur++;
        // Add this index to the result array.
        was.add(cur);
        res[i] = cur;
    }
    return res;
}

它背后的想法是在未使用的值数组中生成一个新数字的位置（此数组未明确维护），然后计算实际索引值，同时考虑已使用的数字。
这种方法有什么好处，它只调用nextInt numberOfIndices次，并且无论nextInt返回什么，都可以保证生成不同的数字。

Answer 2

    int[] integer_array = {0, 1, 2, 3, 4, 5, 6, 7};
    Random r = new Random();

    int random_no = r.nextInt(integer_array.length);
    System.out.println(random_no);

    int random_no2;
    do {
      random_no2 = r.nextInt(integer_array.length);
    } while (random_no2 == random_no);
    System.out.println(random_no2);

    int random_no3;
    do {
      random_no3 = r.nextInt(integer_array.length);
    } while (random_no3 == random_no || random_no3 == random_no2);
    System.out.println(random_no3);

从一组数字中获取非重复随机数

2 个答案: