rust - 如何使用迭代器和范围创建阶乘的非递归计算？

Question

我遇到了一个不断困扰我的 RuSTLings 练习:

pub fn factorial(num: u64) -> u64 {
    // Complete this function to return factorial of num
    // Do not use:
    // - return
    // For extra fun don't use:
    // - imperative style loops (for, while)
    // - additional variables
    // For the most fun don't use:
    // - recursion
    // Execute `rustlings hint iterators4` for hints.
}

解决方案的提示告诉我...

In an imperative language you might write a for loop to iterate through multiply the values into a mutable variable. Or you might write code more functionally with recursion and a match clause. But you can also use ranges and iterators to solve this in rust.

我试过这种方法，但我错过了一些东西:

if num > 1 {
    (2..=num).map(|n| n * ( n - 1 ) ??? ).???
} else {
    1
}

我必须使用类似 .take_while 的东西吗？而不是 if ?

Answer 1

阶乘定义为从起始数到 1 的所有数的乘积。我们使用该定义和 Iterator::product :

fn factorial(num: u64) -> u64 {
    (1..=num).product()
}

如果你看 the implementation的 Product对于整数，您会看到它使用 Iterator::fold 引擎盖下:

impl Product for $a {
    fn product<I: Iterator<Item=Self>>(iter: I) -> Self {
        iter.fold($one, Mul::mul)
    }
}

你可以自己硬编码:

fn factorial(num: u64) -> u64 {
    (1..=num).fold(1, |acc, v| acc * v)
}

也可以看看:

How to sum the values in an array, slice, or Vec in Rust?

How do I sum a vector using fold?