我正在尝试编写remove（T项）和T remove（int index）函数。 remove（T item）删除您输入的节点。 T remove（int index）从您输入的索引中删除节点。我尝试编写删除函数的每一种方式都会出错。有关如何入门的任何想法？
我试着写这个。当我编译它时似乎不起作用。
public boolean remove(T item) {
if(first==null)
System.out.println("The List Is Empty");
else{
Node<T> current = first;
Node<T> previous = current.getPrevious();
while(current!=null && !current.getData().equals(item)){
previous=current;
current=current.getNext();
}
if(current == first && current.getData().equals(item)){
first = first.getNext();
}else if(current != null)
previous.setNext(current.getNext());
}
size--;
return true;
}
这是代码。
public class LinkedList<T> implements Iterable<T> {
private int size = 0;
private Node<T> first = null,
last = null;
class Node<T> {
private T data;
private Node<T> next = null;
private Node<T> previous = null;
public Node(T item) {
data = item;
}
public T getData() { return data; }
public void setData(T data) { this.data = data; }
public Node<T> getNext() { return next; }
public void setNext(Node<T> next) { this.next = next; }
public Node<T> getPrevious() { return previous; }
public void setPrevious(Node<T> previous) { this.previous = previous; }
}
public LinkedList() {
first = new Node<T>(null); // no data in first
last = new Node<T>(null); // no data in last
first.setNext(last);
last.setPrevious(first);
}
public String toString() {
StringBuilder build = new StringBuilder("[");
Node<T> current = first;
while (current.getNext() != last) {
current = current.getNext();
build.append(current.getData());
if (current.getNext() != last)
build.append(", ");
}
return build.toString() + "]";
}
public int size() { return size; }
// add an item at the end of the list
public void add(T item) {
Node<T> newNode = new Node<T>(item);
Node<T> nextToLast = last.getPrevious();
nextToLast.setNext(newNode);
newNode.setPrevious(nextToLast);
newNode.setNext(last);
last.setPrevious(newNode);
size++;
}
// add an item at any position other than the end
// adds such the new item is at position "index" after
// add is done
public void add(int index, T item) {
Node<T> current = first;
Node<T> newNode = new Node<T>(item);
for (int i=0; i<= index; i++)
current = current.getNext();
Node<T> previous = current.getPrevious();
previous.setNext(newNode);
newNode.setPrevious(previous);
current.setPrevious(newNode);
newNode.setNext(current);
size++;
}
public T set(int index, T item) {
if (index >= size) throw new ArrayIndexOutOfBoundsException();
Node<T> current = first.getNext(); // before it was first;
for (int i=0; i<index; i++)
current = current.getNext();
T answer = current.getData();
current.setData(item);
return answer;
}
public boolean remove(T item) {
return true;
}
public T remove(int index) {
return null;
}
public boolean contains(T item) {
return false;
}
public T get(int index) {
return null;
}
// this is the code from the singly linked list implementation,
// and without empty first and last nodes.
public Iterator<T> iterator() {
return new Iterator<T>( ) {
private Node<T> current = null, previous = null;
public boolean hasNext() {
return current != last;
}
public T next() {
previous = current;
if (current == null) current = first;
else current = current.getNext();
return current.getData();
}
public void remove() {
if (current == null)
throw new IllegalStateException();
if (previous == null)
first = current.getNext();
else {
previous.setNext(current.getNext());
if (current == last)
last = previous;
current = previous;
previous = null;
}
size--;
}
};
}
public static void main(String[] args) {
Scanner console = new Scanner(System.in);
LinkedList<Double> theList = new LinkedList<Double>();
int choice = -1;
int index;
do {
System.out.println("The list is " + theList);
System.out.println("Enter a choice between 1 and 9");
System.out.println("1 = add 2 = addindex 3 = set 4 = remove");
System.out.println("5 = removeindex 6 = contains 7 = get 8 = Iterator");
System.out.println("9 = Iterator remove");
choice = console.nextInt();
switch(choice) {
case 1:
System.out.println("Number");
theList.add(console.nextDouble());
break;
case 2:
System.out.println("Index");
index = console.nextInt();
System.out.println("Number");
theList.add(index, console.nextDouble());
break;
case 3:
System.out.println("Index");
index = console.nextInt();
System.out.println("Number");
theList.set(index, console.nextDouble());
break;
case 4:
System.out.println("Number");
theList.remove(console.nextDouble());
break;
case 5:
System.out.println("Index");
index = console.nextInt();
theList.remove(index);
break;
case 6:
System.out.println("Number");
System.out.println(theList.contains(console.nextDouble()));
break;
case 7:
System.out.println("Index");
System.out.println(theList.get(console.nextInt()));
break;
case 8:
System.out.print("Output using Iterator: [");
for (Iterator<Double> it = theList.iterator(); it.hasNext(); ) {
System.out.print(it.next());
if (it.hasNext())
System.out.print(", ");
else
System.out.println("]");
}
break;
case 9:
System.out.println("Index");
index = console.nextInt();
Iterator<Double> i = theList.iterator();
for (int x=0; x<index; x++)
i.next();
i.remove();
System.out.println(theList);
default:
System.out.println("Enter a choice between 1 and 9");
}
} while (choice != -1);
}
}
答案 0 :(得分：0)
在您的构造函数中，您正在初始化first＆amp; last节点包含null个数据。
public LinkedList() {
first = new Node<T>(null); // no data in first
last = new Node<T>(null); // no data in last
first.setNext(last);
last.setPrevious(first);
}
在remove(T item)实现中，您正在检查first是否为null（由于构造函数而不是真的）。
因此代码将进入else块并且在first节点中数据为空，因此您将获得NPE @以下行：
while(current!=null && !current.getData().equals(item)){ if(current == first && current.getData().equals(item)){ previous.setNext(current.getNext()); 您需要通过正确的数据验证来修复constructor（删除不必要的变量初始化）以及remove方法。
仅供参考：您应该为每种方法都有测试用例来捕获这些问题。