如何在一个范围内生成一个随机数但排除一些？ [英] How can I generate a random number within a range but exclude some?

问题描述

如何在一个范围内生成随机数但排除一些，而不继续生成并检查生成的数字是否是我要排除的数字之一？

How can I generate a random number within a range but exclude some, without keep generating and checking if the generated number is one of those that I want to exclude?

推荐答案

每次随机无重生的一种可能解决方案是使用以下算法：

One possible solution without regeneration the random each time is to use the following algorithm:

public int getRandomWithExclusion(Random rnd, int start, int end, int... exclude) {
    int random = start + rnd.nextInt(end - start + 1 - exclude.length);
    for (int ex : exclude) {
        if (random < ex) {
            break;
        }
        random++;
    }
    return random;
}

可以使用数组引用调用此方法，例如

This method can be either called with an array reference, e.g.

int[] ex = { 2, 5, 6 };
val = getRandomWithExclusion(rnd, 1, 10, ex)

或直接插入号码进入通话：

or by directly inserting the numbers into the call:

val = getRandomWithExclusion(rnd, 1, 10, 2, 5, 6)

它在 start 和<$之间生成一个随机数（int） c $ c>结束（包括两者）并且不会为您提供数组 exclude 中包含的任何数字。所有其他数字的概率相等。请注意，必须保留以下约束： exclude 按升序排序，所有数字都在提供的范围内，并且所有数字都是相互不同的。

It generates a random number (int) between start and end (both inclusive) and does not give you any number which is contained in the array exclude. All other numbers occur with equal probability. Note, that the following constrains must hold: exclude is sorted ascendingly and all numbers are within the range provided and all of them are mutually different.

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