linked-list - 在我的 Rust 单链表中实现 .pop() 的更好方法是什么？

Question

我已经在 Rust 中实现了我自己的单向链表版本，作为我学习它的挑战之一，我对除了 .pop() 方法之外的所有东西都很满意。使用 2 个 while 循环非常丑陋且效率低下，但我没有找到其他方法来解决将索引处的节点 len() - 2 设置为无(弹出列表)并使用索引处节点的数据的问题len() - 1 表示 Some(data) 返回值(返回弹出的元素)。

GitHub Link

pub struct SimpleLinkedList<T> {
    head: Option<Box<Node<T>>>,
}

struct Node<T> {
    data: T,
    next: Option<Box<Node<T>>>,
}

impl<T> Default for SimpleLinkedList<T> {
    fn default() -> Self {
        SimpleLinkedList { head: None }
    }
}

impl<T: Copy> Clone for SimpleLinkedList<T> {
    fn clone(&self) -> SimpleLinkedList<T> {
        let mut out: SimpleLinkedList<T> = SimpleLinkedList::new();
        let mut cur = &self.head;
        while let Some(node) = cur {
            cur = &node.next;
            out.push(node.data)
        }
        out
    }
}

impl<T> SimpleLinkedList<T> {
    pub fn new() -> Self {
        Default::default()
    }

    pub fn len(&self) -> usize {
        let mut c = 0;
        let mut cur = &self.head;
        while let Some(node) = cur {
            cur = &node.next;
            c += 1;
        }
        c
    }

    pub fn is_empty(&self) -> bool {
        self.len() == 0
    }

    pub fn push(&mut self, _element: T) {
        let mut cur = &mut self.head;
        match cur {
            Some(_) => {
                while let Some(node) = cur {
                    cur = &mut node.next;
                }
            }
            None => (),
        }
        *cur = Some(Box::from(Node {
            data: _element,
            next: None,
        }));
    }

    pub fn pop(&mut self) -> Option<T>
    where
        T: Copy,
    {
        let length = &self.len();
        let mut cur = &mut self.head;
        let mut out = None;
        match cur {
            Some(_) if *length > 1usize => {
                let mut c = 0usize;
                while let Some(node) = cur {
                    cur = &mut node.next;
                    if c >= length - 1 {
                        out = Some(node.data);
                        break;
                    }
                    c += 1;
                }

                c = 0usize;
                cur = &mut self.head;
                while let Some(node) = cur {
                    cur = &mut node.next;
                    if c == length - 2 {
                        break;
                    }
                    c += 1;
                }
            }
            Some(node) => out = Some(node.data),
            None => (),
        }
        *cur = None;
        out
    }

    pub fn peek(&self) -> Option<&T> {
        let cur = &self.head;
        match cur {
            Some(node) => Some(&node.data),
            None => None,
        }
    }
}

impl<T: Copy> SimpleLinkedList<T> {
    pub fn rev(&self) -> SimpleLinkedList<T> {
        let mut clone = self.clone();
        let mut out: SimpleLinkedList<T> = SimpleLinkedList::new();
        while let Some(val) = clone.pop() {
            out.push(val)
        }
        out
    }
}

impl<'a, T: Copy> From<&'a [T]> for SimpleLinkedList<T> {
    fn from(_item: &[T]) -> Self {
        let mut out: SimpleLinkedList<T> = SimpleLinkedList::new();
        for &e in _item.iter() {
            out.push(e);
        }
        out
    }
}

impl<T> Into<Vec<T>> for SimpleLinkedList<T> {
    fn into(self) -> Vec<T> {
        let mut out: Vec<T> = Vec::new();
        let mut cur = self.head;
        while let Some(node) = cur {
            cur = node.next;
            out.push(node.data)
        }
        out
    }
}

Answer 1

您可以通过跟踪您看到的最后一个元素(然后在最后更新它)来避免重新遍历列表。

如果你对自己的做法太天真了，你会遇到麻烦；您的“上一个”指针保留列表其余部分的所有权，借用检查器不允许这样做。诀窍是随时断开该链接 - 为此，您可以使用 mem::replace 函数。完成此操作后，您必须在再次失去对前一个节点的跟踪之前将其放回原位。

这是完整的样子(你必须原谅我对 unwrap 的自由使用——我确实认为它让事情变得更清晰):

pub fn pop(&mut self) -> Option<T>
    where T : Copy,
{
    use std::mem::replace;

    let curr = replace(&mut self.head, None);

    if curr.is_none() { // list started off empty; nothing to pop
        return None;
    }

    let mut curr = curr.unwrap(); // safe because of the check above

    if let None = curr.next { // popped the last element
        return Some(curr.data);
    }

    let mut prev_next = &mut self.head;

    while curr.next.is_some() {
        // Take ownership of the next element
        let nnext = replace(&mut curr.next, None).unwrap();

        // Update the previous element's "next" field
        *prev_next = Some(curr);

        // Progress to the next element
        curr = nnext;

        // Progress our pointer to the previous element's "next" field
        prev_next = &mut prev_next.as_mut().unwrap().next;

    }

    return Some(curr.data);
}

顺便说一句，如果您愿意稍微更改接口(interface)以便我们每次都返回一个"new"列表(在 pop 函数中拥有所有权)，那么所有这些指针改组都会简化很多)，或使用持久数据结构，就像他们在 Learning Rust with entirely too many linked lists 中所做的那样(已在评论中提及):

pub fn pop_replace(self) -> (Option<T>, Self) {
    // freely mutate self and all the nodes
}

你会这样使用:

let elem, list = list.pop();