这是leetcode 19题，删除链表倒数第n个节点。

我用的双指针，找到要删除节点的前继sec，现在的问题是，我想把&，改成&mut，进行链表节点的删除操作，

想请教下大家，这一步有什么方法嘛

pub fn remove_nth_from_end(head: Option<Box<ListNode>>, n: i32) -> Option<Box<ListNode>> {
    let mut dummy = Some(Box::new(ListNode{val: 0, next: head}));

    let mut fst = &dummy;
    let mut sec = &dummy;

    for _ in 0..n+1{
        fst = &fst.as_ref()?.next;
    }
    loop {
        if fst.is_none() {
            break;
        }
        fst = &fst.as_ref()?.next;
        sec = &sec.as_ref()?.next;
    }

    // now, just remove sec.next, but i can't change sec
    
    dummy?.next
}