用Java方法计算第n个根 [英] Calculating nth root in Java using power method

问题描述

我试图使用 Math.pow（n，1.0 / 3）在java中获得立方根，但是由于它分双倍，它不会返回确切的回答。例如，125，这给了4.9999999999。有没有解决这个问题？我知道有一个立方根函数，但是我想解决这个问题，所以我可以计算出更高的根。

I was trying to get a cubic root in java using Math.pow(n, 1.0/3) but because it divides doubles, it doesn't return the exact answer. For example, with 125, this gives 4.9999999999. Is there a work-around for this? I know there is a cubic root function but I'd like to fix this so I can calculate higher roots.

我不想圆，因为我想知道是否一个数字通过执行以下操作来实现整数： Math.pow（n，1.0 / 3）％（（int）Math.pow（n，1.0 / 3））。

I would not like to round because I want to know whether a number has an integer root by doing something like this: Math.pow(n, 1.0 / 3) % ((int) Math.pow(n, 1.0 / 3)).

推荐答案

由于不可能使用 double ，你有三个选择：

Since it is not possible to have arbitrary-precision calculus with double, you have three choices:

1. 定义一个精度，你决定一个 double value是一个整数。


2. 测试您的 double 的舍入值是否正确。


3. 在 BigDecimal 对象，它支持任意精度的双重值。

1. Define a precision for which you decide whether a double value is an integer or not.
2. Test whether the rounded value of the double you have is a correct result.
3. Do calculus on a BigDecimal object, which supports arbitrary-precision double values.

选项1

Option 1

private static boolean isNthRoot(int value, int n, double precision) {
    double a = Math.pow(value, 1.0 / n);
    return Math.abs(a - Math.round(a)) < precision; // if a and round(a) are "close enough" then we're good
}

这种方法的问题是如何定义足够近。这是一个主观问题，这取决于你的要求。

The problem with this approach is how to define "close enough". This is a subjective question and it depends on your requirements.

private static boolean isNthRoot(int value, int n) {
    double a = Math.pow(value, 1.0 / n);
    return Math.pow(Math.round(a), n) == value;
}

此方法的优点是无需定义精度。但是，我们需要执行另一个 pow 操作，这样会影响性能。

The advantage of this method is that there is no need to define a precision. However, we need to perform another pow operation so this will affect performance.

没有内置的方法来计算BigDecimal的双倍幂。 此问题将为您提供有关如何做到这一点的见解。

There is no built-in method to calculate a double power of a BigDecimal. This question will give you insight on how to do it.

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