[英]Choose random integer in a range bigger than Integer.MAX_VALUE?
I know there are another questions about random in a range but none of their answers accomplishes what I am trying to do. 我知道在一个范围内还有另外一个关于随机的问题,但他们的答案都没有完成我想要做的事情。 Actually they have the same error I have.
实际上他们有同样的错误。 I wrote this simple function to generate random with range.
我写了这个简单的函数来生成随机范围。
Random m_random = new Random();
...
public int RandomWithRange(int min, int max) {
return m_random.nextInt(max - min + 1) + min;
}
If range is bigger than Integer.MAX_VALUE, it throws an IllegalArgumentException: n must be positive. 如果range大于Integer.MAX_VALUE,则抛出IllegalArgumentException:n必须为正数。 I know it overflows and turn to a negative number.
我知道它溢出并转向负数。 My question is how to handle that?
我的问题是如何处理?
Example ranges; 示例范围;
Note: min and max must be inclusive. 注意:min和max必须包含在内。
You cannot use int in this case. 在这种情况下,您不能使用int。 You need to go with BigInteger .
你需要使用BigInteger 。 The following constructor does what you want (need some tweaking for your needs of course):
以下构造函数执行您想要的操作(当然需要根据您的需要进行调整):
BigInteger(int numBits, Random rnd)
Constructs a randomly generated BigInteger, uniformly distributed over the range 0 to (2numBits - 1), inclusive.
构造一个随机生成的BigInteger,均匀分布在0到(2numBits - 1)的范围内,包括0和(2numBits - 1)。
The problem you have is that (max - min)
overflows and gives you a negative value. 你遇到的问题是
(max - min)
溢出并给你一个负值。
You can use a long
instead. 你可以使用
long
而不是。
public int randomWithRange(int min, int max) {
return (int) ((m_random.nextLong() & Long.MAX_VALUE) % (1L + max - min)) + min;
}
你有没有考虑过随机加倍然后再回到int
return (int)(m_random.nextDouble() * ((double)max - (double)min) + min);
The most "dumb but definitely correct" solution I can think of: 我能想到的最“愚蠢但绝对正确”的解决方案:
if (max - min + 1 > 0) // no overflow
return delta + random.nextInt(max - min + 1);
else {
int result;
do {
result = random.nextInt();
} while (result < min || result > max);
// finishes in <= 2 iterations on average
return result;
}
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