java – 从1-50的生成器生成1-100的随机数

Print random numbers from 1-100 using the given getrnd50() method
which generates the random numbers from 1-50. Each random number
should be printed only once and in random order. Use of no other random number generator
is allowed and i was not allowed to change the definition of
getrnd50().

我想出了以下代码,它给出了正确的输出.

import java.util.Random;

public class Test {

public static void main(String[] args) {
    int[] rs = new int[100];
    int count = 0;
    int k;
    while (count != 100) {

        // I decided to simply multiply the result of `getrnd50()` by 2. 
        // But since that would generate only even numbers,k = getrnd50() * 2;

        // I decided to randomly subtract 1 from the numbers. 
        // Which i accomlished as follows.          

        if (getrnd50() <= 25) // 25 is to half the possibilities.
            k--;

        // Every number is to be stored in its own unique location 
        // in the array `rs`,the location is `(number-1)`. 
        // Every time a number is generated it is checked whether it has 
        // already been generated. If not,it is saved in its position,printed and 
        // `count` is incremented.

        if (rs[k-1] == 0) {
            rs[k-1] = k;
            count++;
            System.out.print(k + " ");
        }
    }
}
// This is the method and i am not supposed to touch it.
static int getrnd50() {
    Random rand = new Random();
    return (1 + rand.nextInt(50));
}

}

虽然它在那一轮被接受,但在下一轮中,面试官告诉我getrnd50()是一种昂贵的方法,即使在最好的情况下,我必须为每个生成的数字调用它两次.即1-100的200次.在最坏的情况下,它将是无限的,平均情况下数万.他要求我优化代码,以便显着改善平均情况.

当我表示无法做到时,他给了我一个暗示,他说：

To consider the number of numbers generated while generating a new
number. For ex. if count becomes 99 i don’t have to call getrnd50() I
can simply find the remaining number and print it.

虽然我理解他的漂移我不知道它会如何帮助我,所以显然我被拒绝了.现在我很想知道答案.帮我！ Thanx提前！

注意：如果有人懒得写一个冗长的代码只是指出数字生成部分,其余的很容易.此外,我们不一定遵循提示.