我理解LinkedList的概念，以及单/双链表。但是，我不明白如何将其实现到我的代码中？
我必须将这个单独的列表转换为双向链表：
public class MyLinkedList<E> extends MyAbstractList<E> {
private Node<E> head, tail;
/** Create a default list */
public MyLinkedList() {
}
/** Create a list from an array of objects */
public MyLinkedList(E[] objects) {
super(objects);
}
/** Return the head element in the list */
public E getFirst() {
if (size == 0) {
return null;
}
else {
return head.element;
}
}
/** Return the last element in the list */
public E getLast() {
if (size == 0) {
return null;
}
else {
return tail.element;
}
}
/** Add an element to the beginning of the list */
public void addFirst(E e) {
Node<E> newNode = new Node<E>(e); // Create a new node
newNode.next = head; // link the new node with the head
head = newNode; // head points to the new node
size++; // Increase list size
if (tail == null) // the new node is the only node in list
tail = head;
}
/** Add an element to the end of the list */
public void addLast(E e) {
Node<E> newNode = new Node<E>(e); // Create a new for element e
if (tail == null) {
head = tail = newNode; // The new node is the only node in list
}
else {
tail.next = newNode; // Link the new with the last node
tail = tail.next; // tail now points to the last node
}
size++; // Increase size
}
@Override /** Add a new element at the specified index
* in this list. The index of the head element is 0 */
public void add(int index, E e) {
if (index == 0) {
addFirst(e);
}
else if (index >= size) {
addLast(e);
}
else {
Node<E> current = head;
for (int i = 1; i < index; i++) {
current = current.next;
}
Node<E> temp = current.next;
current.next = new Node<E>(e);
(current.next).next = temp;
size++;
}
}
/** Remove the head node and
* return the object that is contained in the removed node. */
public E removeFirst() {
if (size == 0) {
return null;
}
else {
Node<E> temp = head;
head = head.next;
size--;
if (head == null) {
tail = null;
}
return temp.element;
}
}
/** Remove the last node and
* return the object that is contained in the removed node. */
public E removeLast() {
if (size == 0) {
return null;
}
else if (size == 1) {
Node<E> temp = head;
head = tail = null;
size = 0;
return temp.element;
}
else {
Node<E> current = head;
for (int i = 0; i < size - 2; i++) {
current = current.next;
}
Node<E> temp = tail;
tail = current;
tail.next = null;
size--;
return temp.element;
}
}
@Override /** Remove the element at the specified position in this
* list. Return the element that was removed from the list. */
public E remove(int index) {
if (index < 0 || index >= size) {
return null;
}
else if (index == 0) {
return removeFirst();
}
else if (index == size - 1) {
return removeLast();
}
else {
Node<E> previous = head;
for (int i = 1; i < index; i++) {
previous = previous.next;
}
Node<E> current = previous.next;
previous.next = current.next;
size--;
return current.element;
}
}
@Override /** Override toString() to return elements in the list */
public String toString() {
StringBuilder result = new StringBuilder("[");
Node<E> current = head;
for (int i = 0; i < size; i++) {
result.append(current.element);
current = current.next;
if (current != null) {
result.append(", "); // Separate two elements with a comma
}
else {
result.append("]"); // Insert the closing ] in the string
}
}
return result.toString();
}
@Override /** Clear the list */
public void clear() {
size = 0;
head = tail = null;
}
@Override /** Return true if this list contains the element e */
public boolean contains(E e) {
System.out.println("Implementation left as an exercise");
return true;
}
@Override /** Return the element at the specified index */
public E get(int index) {
System.out.println("Implementation left as an exercise");
return null;
}
@Override /** Return the index of the head matching element in
* this list. Return -1 if no match. */
public int indexOf(E e) {
System.out.println("Implementation left as an exercise");
return 0;
}
@Override /** Return the index of the last matching element in
* this list. Return -1 if no match. */
public int lastIndexOf(E e) {
System.out.println("Implementation left as an exercise");
return 0;
}
@Override /** Replace the element at the specified position
* in this list with the specified element. */
public E set(int index, E e) {
System.out.println("Implementation left as an exercise");
return null;
}
@Override /** Override iterator() defined in Iterable */
public java.util.Iterator<E> iterator() {
return new LinkedListIterator();
}
private void checkIndex(int index) {
if (index < 0 || index >= size)
throw new IndexOutOfBoundsException
("Index: " + index + ", Size: " + size);
}
private class LinkedListIterator
implements java.util.Iterator<E> {
private Node<E> current = head; // Current index
@Override
public boolean hasNext() {
return (current != null);
}
@Override
public E next() {
E e = current.element;
current = current.next;
return e;
}
@Override
public void remove() {
System.out.println("Implementation left as an exercise");
}
}
private static class Node<E> {
E element;
Node<E> next;
public Node(E element) {
this.element = element;
}
}
}
所以我理解添加一个前一个指针，我必须这样做：
private static class Node<E> {
E element;
Node<E> next;
Node<E> previous;
public Node(E element) {
this.element = element;
}
现在，我很困惑 - 我如何实际为双向链表实现必要的方法？我不是要求某人完成整个任务，但有人能告诉我一个解决方法的例子吗？
我也无法理解，当我Node<E> previous时，程序如何知道我引用了前一个元素？
答案 0 :(得分：2)
我也不明白，只是因为我做“Node previous”，程序如何知道我引用了前一个元素？
程序 知道您正在引用前一个元素。您，程序员，需要维护（保持最新和正确）该引用。
答案 1 :(得分：1)
每个节点也会引用前一个节点，因为它引用了下一个节点。
更改Node的数据结构以向前一节点添加额外的Node引用。所以，现在当有人调用add(E e)方法时，您会在尾部添加该元素并设置先前的e 引用（这将是新的尾节点） 到较旧的尾节点（调用add之前的尾部）。
因此，通过这种方式，您可以管理对前一个和下一个节点的两个引用
还要为双链接列表类pervious()和相关链接添加新方法。