我必须为我的项目实现一个单链列表,但我无法使remove方法起作用。我在这里搜索了答案,但是找不到包含尾部引用的任何答案。我的项目需要在列表中有头和尾参考,并且需要在需要时进行更新。这是我的课程和remove方法:
public class BasicLinkedList<T> implements Iterable<T> {
public int size;
protected class Node {
protected T data;
protected Node next;
protected Node(T data) {
this.data = data;
next = null;
}
}
protected Node head;
protected Node tail;
public BasicLinkedList() {
head = tail = null;
}
public BasicLinkedList<T> addToEnd(T data) {
Node n = new Node(data);
Node curr = head;
//Check to see if the list is empty
if (head == null) {
head = n;
tail = head;
} else {
while (curr.next != null) {
curr = curr.next;
}
curr.next = n;
tail = n;
}
size++;
return this;
}
public BasicLinkedList<T> addToFront(T data) {
Node n = new Node(data);
if(head == null){
head = n;
tail = n;
}
n.next = head;
head = n;
size++;
return this;
}
public T getFirst() {
if (head == null) {
return null;
}
return head.data;
}
public T getLast() {
if(tail == null){
return null;
}
return tail.data;
}
public int getSize() {
return size;
}
public T retrieveFirstElement() {
// Check to see if the list is empty
if (head == null) {
return null;
}
Node firstElement = head;
Node curr = head.next;
head = curr;
size--;
return firstElement.data;
}
public T retrieveLastElement() {
Node curr = head;
Node prev = head;
// Check to see if the list is empty
if (head == null) {
return null;
} else {
// If there's only one element in the list
if (head.next == null) {
curr = head;
head = null;
} else {
while (curr.next != null) {
prev = curr;
curr = curr.next;
}
tail = prev;
tail.next = null;
}
}
size--;
return curr.data;
}
public void remove(T targetData, Comparator<T> comparator) {
Node prev = null, curr = head;
while (curr != null) {
if (comparator.compare(curr.data, targetData) == 0) {
//Check to see if we need to remove the very first element
if (curr == head) {
head = head.next;
curr = head;
}
//Check to see if we need to remove the last element, in which case update the tail
else if(curr == tail){
curr = null;
tail = prev;
prev.next = null;
}
//If anywhere else in the list
else {
prev.next = curr.next;
curr = curr.next;
}
size--;
} else {
prev = curr;
curr = curr.next;
}
}
}
public Iterator<T> iterator() {
return new Iterator<T>() {
Node current = head;
@Override
public boolean hasNext() {
return current != null;
}
@Override
public T next() {
if(hasNext()){
T data = current.data;
current = current.next;
return data;
}
return null;
}
@Override
public void remove(){
throw new UnsupportedOperationException("Remove not implemented.");
}
};
}
}
我已经经历了这种方法的许多迭代,每次我丢失头参考,尾参考或者我不删除该元素时,都会为解决这个问题而烦恼。作为参考,这里是我正在运行的测试。我什至没有通过测试,只是说失败跟踪。
public void testRemove(){
BasicLinkedList<String> basicList = new BasicLinkedList<String>();
basicList.addToEnd("Blue");
basicList.addToEnd("Red");
basicList.addToEnd("Magenta");
//Blue -> Red -> Magenta -> null
basicList.remove("Red", String.CASE_INSENSITIVE_ORDER);
//Blue -> Magenta -> null
assertTrue(basicList.getFirst().equals("Blue"));
//getLast() returns the tail node
assertTrue(basicList.getLast().equals("Magenta"));
}
编辑:忘记提及remove方法应该从列表中删除目标数据的所有实例。
我只看到1个错误。如果您的列表最初是空的,则以下方法将导致一个循环,在该循环中您有一个节点,下一个节点指向自身:
public BasicLinkedList<T> addToFront(T data) {
Node n = new Node(data);
// The list was empty so this if is true
if(head == null){
head = n;
tail = n;
}
n.next = head;
// now head == n and n.next == head == n so you've got a circle
head = n;
size++;
return this;
}
您可以这样解决:
public BasicLinkedList<T> addToFront(T data) {
Node n = new Node(data);
if(head == null){
tail = n;
}
n.next = head;
head = n;
size++;
return this;
}
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