java - LinkedList addAfter方法不会更新下一个节点的前一个成员变量

Question

我正在尝试构建一个具有 addAfter 方法的链接列表。由于某种原因，add方法会插入一个新节点，但它不会更新下一个节点的前一个成员变量。这是我的输出。请注意，在第二个列表中，“先前”输出之一不正确。如何修复我的 addafter 方法？

public class DoubleLinkedList<E> implements List211<E> {
    private static class DLinkedNode<E> {
        private E data;
        private DLinkedNode<E> next = null;
        private DLinkedNode<E> prev = null;

        private DLinkedNode(E dataItem) {
            data = dataItem;
        }

        private DLinkedNode(E dataItem, DLinkedNode<E> nextNodeRef,
                DLinkedNode<E> prevNodeRef) {
            data = dataItem;
            next = nextNodeRef;
            prev = prevNodeRef;
        }

        public String toString() {
            return data.toString();
        }

        public void setPrev(DLinkedNode<E> prev) {
            this.prev = prev;
        }
    }

    private DLinkedNode<E> head = null;
    private DLinkedNode<E> tail = null;
    private int size;

    // http://codereview.stackexchange.com/questions/63171/implementation-of-a-doubly-linked-list-in-java
    public void addFirst(E item) {

        DLinkedNode<E> newNode = new DLinkedNode<E>(item);
        if (size < 1) {
            newNode.next = null;
            newNode.prev = null;
            head = newNode;
            tail = newNode;

        } else {
            head.prev = newNode;
            newNode.next = head;
            newNode.prev = null;
            head = newNode;
        }
        size++;
    }


    private void addAfter(DLinkedNode<E> node, E item) {
        //WHAT AM I DOING WRONG
        DLinkedNode<E> newNode = new DLinkedNode<E>(item, node.next, node);
        node.next = newNode;
        //node.next.next = newNode; (maybe?)
        if (node == tail) { 
          tail = newNode;
        }
        size++;
    }

    private E removeAfter(DLinkedNode<E> node) {
        DLinkedNode<E> tempNext = node.next;
        if (tempNext != null) {
            node.next = tempNext.next;
            node.next.prev = node;
            size--;
            return tempNext.data;
        } else {
            return null;
        }
    }

    private E removeFirst() {
        DLinkedNode<E> temp = head;
        if (head != null) {
            head = head.next;
            head.prev = null;
        }
        if (temp != null) {
            size--;
            return temp.data;
        } else {
            return null;
        }
    }

    public String toString() {
        DLinkedNode<String> nodeRef = (DLinkedNode<String>) head;
        StringBuilder result = new StringBuilder();
        while (nodeRef != null) {
            result.append(nodeRef.data);
            if (nodeRef.next != null) {
                result.append(" ==> ");
            }
            nodeRef = nodeRef.next;
        }
        return result.toString();
    }

    private DLinkedNode<E> getNode(int index) {
        DLinkedNode<E> node = head;
        for (int i = 0; i < index && node != null; i++) {
            node = node.next;
        }
        return node;
    }

    public E get(int index) {
        checkBounds(index);
        DLinkedNode<E> node = getNode(index);
        return node.data;
    }

    public E set(int index, E newValue) {
        DLinkedNode<E> node = getNode(index);
        E result = node.data;
        node.data = newValue;
        return result;
    }

    private void checkBounds(int index) {
        if (index < 0 || index > size) {
            throw new IndexOutOfBoundsException(Integer.toString(index));
        }
    }

    public void add(int index, E item) {
        checkBounds(index);
        if (index == 0) {
            addFirst(item);

        } else {
            DLinkedNode<E> node = getNode(index - 1);
            addAfter(node, item);

        }
    }

    public boolean add(E item) {
        add(size, item);
        return true;
    }

    @Override
    public E remove(int index) {
        checkBounds(index);
        if (index == 0) {
            this.removeFirst();
        } else {
            DLinkedNode<E> myNode = getNode(index - 1);
            return removeAfter(myNode);
        }

        return null;
    }

    @Override
    public int size() {
        return size;
    }

    public void printLinkedList() {
        System.out.print(this.getClass().getSimpleName() + ": ");
        DLinkedNode<E> myNode = head;
        for (int i = 0; i < size && myNode != null; i++) {
            if (i == size - 1) {
                System.out.print(myNode.toString() + " [next: " + myNode.next
                        + ", previous:" + myNode.prev + "] ");
            } else {
                System.out.print(myNode.toString() + " [next: " + myNode.next
                        + ", previous:" + myNode.prev + "] " + ", ");
            }
            myNode = myNode.next;
        }
    }
}

这是我的主要方法:

public class MainTester {    
    public static void main(String[] args) {
        DoubleLinkedList myList = new DoubleLinkedList();
        double one = 1.0;
        double two = 2.0;
        double three = 3.0;
        double four = 4.0;
        double five = 5.0;
        double six = 6.0;
        myList.addFirst(one);
        myList.add(two);
        myList.add(three);
        myList.add(four);
        myList.add(five);
        myList.add(six);
        myList.printLinkedList();           

        System.out.println("\n\n");
        myList.add(1,45.0);
        myList.printLinkedList();

        /*
        System.out.println("\n\n");
        myList.add(2, three);
        myList.printLinkedList();
        */
    }
}

这是我的输出:

DoubleLinkedList: 1.0 [next: 2.0, previous:null] , 2.0 [next: 3.0, previous:1.0] , 3.0 [next: 4.0, previous:2.0] , 4.0 [next: 5.0, previous:3.0] , 5.0 [next: 6.0, previous:4.0] , 6.0 [next: null, previous:5.0]

DoubleLinkedList: 1.0 [next: 45.0, previous:null] , 45.0 [next: 2.0, previous:1.0] , 2.0 [next: 3.0, previous:1.0] , 3.0 [next: 4.0, previous:2.0] , 4.0 [next: 5.0, previous:3.0] , 5.0 [next: 6.0, previous:4.0] , 6.0 [next: null, previous:5.0]

Answer 1

DLinkedNode<E> newNode = new DLinkedNode<E>(item, node.next, node);
if (node.next != null)
   node.next.prev = newNode;
newNode.next = node.next;
node.next = newNode;
newNode.prev = node;

这应该得到所需的顺序。尝试绘图/可视化 - 有助于对陈述进行排序。