我已经在Rust中实现了我自己版本的单链表,这是我学习它的挑战之一,除了.pop()方法之外,我对我所拥有的一切感到满意。使用2 while循环是非常丑陋和低效的,但我发现没有其他方法可以解决在索引len() - 2设置节点的问题 - 无(弹出列表),并使用索引节点的数据len() - 1表示Some(data)返回值(返回弹出的元素)。
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
}
}
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。
如果你对如何做到这一点很天真,你就会遇到麻烦;您的“上一个”指针保留列表其余部分的所有权,借用检查程序将不允许这样做。诀窍是在你去的时候打破这个链接 - 为此你可以使用mem::replace
函数。完成后,必须先将其恢复,然后再次丢失对上一个节点的跟踪。
这就是完全看起来的样子(你必须原谅我对unwrap
的自由使用 - 我认为它让事情更清晰):
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);
}
顺便说一下,如果你愿意稍微更改一下接口以便我们每次返回一个“新”列表(在pop
函数中获取所有权),或者使用持久数据结构,所有这个指针改组都会简化很多。在Learning Rust with entirely too many linked lists做的(已经在评论中提到):
pub fn pop_replace(self) -> (Option<T>, Self) {
// freely mutate self and all the nodes
}
您将使用哪个:
let elem, list = list.pop();