我面临的问题是弄清楚如何编写方法
public SingleLinkedList copy(Node <E> node) {
}
返回列表的副本。我试过了:
public SingleLinkedList copy(Node <E> node) {
SingleLinkedList<E> temp = new SingleLinkedList<E>();
Node<E> ref = head;
for(Node<E> n = ref ;ref!= null; n = n.next){
temp.add(n, ref.data);
ref = ref.next;
}
return temp;
}
我创建了一个名为temp的新列表,将head更改为ref,遍历列表并将其添加到新列表并返回新列表,但temp.add(n, ref.data)
出现错误。
我可能做错了什么?
class SingleLinkedList<E> {
private static class Node<E> {
private E data;//removed final * private final E data
private Node<E> next;
private Node(E item) {
data = item;
}
}
private Node<E> head;
private int size;
/* Insert item at index, returns true if add is successful. */
public boolean add(int index, E item) {
if (index < 0 || index > size) {
throw new IndexOutOfBoundsException("" + index);
}
if (index == 0) { // adding to the front
Node<E> t = head;
head = new Node<>(item);
head.next = t;
} else { // adding anywhere other than front
Node<E> left = getNode(index - 1);
Node<E> node = new Node(item);
Node<E> right = left.next;
left.next = node;
node.next = right;
}
size++;
return true;
}
/* Add item at end of list, returns true if successful. */
public boolean add(E item) {
return add(size, item);
}
/* Return item at index */
public E get(int index) {
if (index < 0 || index >= size) {
throw new IndexOutOfBoundsException("" + index);
}
return getNode(index).data;
}
/* Return the number of items */
public int size() {
return size;
}
/* Returns a string representation of the list */
@Override
public String toString() {
StringBuilder sb = new StringBuilder();
sb.append("[ ");
for (Node<E> n = head; n != null; n = n.next) {
sb.append(n.data);
sb.append(" ");
}
sb.append("]");
return sb.toString();
}
/* Return the node at location index */
private Node<E> getNode(int index) {
Node<E> n = head;
for (int i = 0; i < index; i++)
n = n.next;
return n;
}
答案 0 :(得分:0)
问题是您需要将该位置作为int传递。我也删除了Node n因为你不需要它。 我认为这应该有用。
public SingleLinkedList copy() {
SingleLinkedList<E> temp = new SingleLinkedList<E>();
int i = 0;
for(Node<E> ref = head ;ref!= null; ref = ref.next){
temp.add(i++, ref.data);
}
return temp;
}
编辑:我忘记删除你根本不需要它的参数。