實現雙向鏈表

Question

我在這個論壇上環顧四周關於雙向鏈表的實現，但我無法理解下面的代碼。

// instance variables of the DoublyLinkedList
    private final Node<E> header;     // header sentinel
    private final Node<E> trailer;    // trailer sentinel
    private int size = 0;       // number of elements in the list
    private int modCount = 0;   // number of modifications to the list (adds or removes)

    /**
     * Creates both elements which act as sentinels
     */
    public DoublyLinkedList() {

        header = new Node<>(null, null, null);      // create header
        trailer = new Node<>(null, header, null);   // trailer is preceded by header
        header.setNext(trailer);                    // header is followed by trailer
    }

我看過關於鏈表和雙鏈表的視頻，但我還沒有看到這種實現。 背后的邏輯是什么，例如： trailer = new Node<>(null, header, null) ？

Answer 1

你可能有一些 DoubleLinkedList 像：

     /**
     * A double linked list.
     *
     */
    public class DoubleLinkedList<E> {
    
        private final Node<E> header;     // header sentinel
        private final Node<E> trailer;    // trailer sentinel
        private int size = 0;       // number of elements in the list
        private int modCount = 0;   // number of modifications to the list (adds or removes)
    
        public DoubleLinkedList() {
            this.header = new Node<>(
                        // The successor of the header is the trailer.
                        // It will be set with: header.setNext(trailer);
                    null,
                        // The predecessor of the header is always null,
                        // because there there is no node before the first
                    null,
                        // The payload of the node is null.
                        // I guess it is just a part of the example.
                    null
            );
    
            this.trailer = new Node<>(
                    // The successor of the trailer is always null,
                    // because there there is no node after the last
                    null,
                    // The predecessor of the trailer is the header
                    // at construction of this object
                    header,
                    // The payload of the node is null.
                    // I guess it is just a part of the example.
                    null
            );
            // Now is the successor of the header set to the trailer.
            header.setNext(trailer);
        }
    
        // Some list methods like add, remove, get, ...
    
    
        /**
         * The nodes of the List
         *
         * @param <T> The type of the stored objects in the list.
         */
        static class Node<T> {
    
            /**
             * The predecessor of this node.
             */
            private Node<T> predecessor;
    
            /**
             * The successor of this node.
             */
            private Node<T> successor;
    
            /**
             * The payload
             */
            private final T payload;
    
            public Node(final Node<T> successor, final Node<T> predecessor, final T payload) {
                this.predecessor = successor;
                this.successor = successor;
                this.payload = payload;
            }
    
            // Getter and Setter:
    
            private Node<T> getPredecessor() {
                return this.predecessor;
            }
    
            private void setNext(final Node<T> next) {
                this.predecessor = next;
            }
    
            private Node<T> getSuccessor() {
                return this.successor;
            }
    
            private void setPrevious(final Node<T> previous) {
                this.successor = previous;
            }
    
            private T getPayload() {
                return this.payload;
            }
        }
    }

這是建築不是很漂亮，但我認為這個解釋符合你的情況。

Answer 2

給定一個列表（任何類型的），您至少需要知道如何到達第一個元素，以及如何判斷何時看到了最后一個元素。

有幾種方法可以滿足這些要求。

對於鏈表，要知道列表從哪里開始，您可能有對第一個節點的簡單引用，或者您可能有一個始終存在的完整“虛擬”節點。

要知道列表在哪里結束，您可能有一個空的“下一個”引用，或者您可能有一個始終存在的完整“虛擬”節點。

虛擬節點方法通常可以產生更清晰的代碼，因為所有實際節點將始終具有“上一個”節點，而所有實際節點將始終具有“下一個”節點。

這似乎是您的代碼提取中采用的方法。