A singly linked list can simply create from tail. But can't from head, I tried many time, code here: https://gist.github.com/tioover/8d7585105c06e01678a8.
In fact, I want search then delete a node in linked list. but I can't traversal linked list with mutable borrow pointer: https://gist.github.com/tioover/526715ed05342ef5b4f1. I tried many time too.
Here's some code from an answer to a similar question. It shows a way of having a list that you can add to the beginning and the end, and the middle for good measure.
#[derive(Debug)]
struct Node<T> {
v: T,
next: Option<Box<Node<T>>>,
}
impl<T> Node<T> {
fn new(v: T) -> Node<T> { Node { v: v, next: None } }
fn push_front(self, head: T) -> Node<T> {
Node {
v: head,
next: Some(Box::new(self)),
}
}
fn push_back(&mut self, tail: T) {
match self.next {
Some(ref mut next) => next.push_back(tail),
None => self.next = Some(Box::new(Node::new(tail))),
}
}
fn push_after(&mut self, v: T) {
let old_next = self.next.take();
let new_next = Node {
v: v,
next: old_next,
};
self.next = Some(Box::new(new_next));
}
}
fn main() {
let mut n = Node::new(2u8);
n.push_back(3u8);
let mut n = n.push_front(0u8);
n.push_after(1u8);
println!("{:?}", n);
}
The important thing is that when we add to the head, we consume the old head by taking it as self. That allows us to move it into a Box which will be the follower of the new head. Removing an item is a straight-forward extension of this example, but you'll need to look forward a bit and handle more edge cases (like what to do if there isn't a second successor).
