remove a node from the end of a linked list in rust

To remove a node from the end of a linked list in Rust, we need to iterate through the entire linked list until we reach the last node. Once we have reached the last node, we can remove it by changing the next pointer of the second-to-last node to None. Here's some sample code that demonstrates this approach:


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

impl<T> Node<T> {
    fn new(data: T) -> Node<T> {
        Node { data, next: None }
    }
}

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

impl<T> LinkedList<T> {
    fn remove_last(&mut self) -> Option<T> {
        let mut current = &mut self.head;

        // Get a reference to the last node and its parent
        let mut last_node: Option<&mut Box<Node<T>>> = None;
        while let Some(ref mut node) = *current {
            if node.next.is_none() {
                last_node = Some(current.as_mut().unwrap());
            }
            current = &mut node.next;
        }

        // Remove the last node
        if let Some(last_node) = last_node {
            let last_data = last_node.data.clone();
            last_node.next = None;
            Some(last_data)
        } else {
            None
        }
    }
}
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In this example, we define a Node struct and a LinkedList struct. The Node struct has a data field and a next field, which is an Option<Box<Node<T>>> because the next node may or may not exist. The LinkedList struct has a single field head which is an Option<Box<Node<T>>>, representing the first node in the linked list.

The remove_last method of the LinkedList struct iterates through the linked list to find the last node and its parent. Once the last node is found, its next field is set to None, effectively removing it from the linked list. The method returns the data stored in the last node, if it exists.

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