我在如何启动此方法时遇到问题。我试图在我的代码中使用递归创建一个删除方法。基本上我有一个公共和私人删除方法。 remove(int)方法是公共的,应该删除列表中指定索引处的元素。我需要解决列表为空和/或删除的元素是列表中的第一个的情况。如果index参数无效,则应抛出IndexOutOfBoundsException。为了允许递归实现,此方法应解决特殊情况并委托删除(int,int,Node)以进行递归。
这是班级:
public class SortedLinkedList<E extends Comparable<E>>
{
private Node first;
private int size;
// ...
}
这是代码:
public void remove(int index)
{
if(index < 0 || index > size)
{
throw new IndexOutOfBoundsException();
}
remove(index++, 0, first);
if (index == 0)
{
if(size == 1)
{
first = null;
}
else
{
first = first.next;
}
}
size--;
}
和私人方法:
private void remove(int index, int currentIndex, Node n)
{
if(index == currentIndex)
{
remove(index, currentIndex, n.next);
}
remove(index, currentIndex, n.next.next);
}
私人课程:
private class Node
{
private E data;
private Node next;
public Node(E data, Node next)
{
this.data = data;
this.next = next;
}
}
void
private void remove(int index, int current, Node n) {
if (n == null || index <= 0 || (index == 1 && n.next == null) {
throw new IndexOutOfBoundsException();
}
if (current == index - 1) {
// Remove 'n.next'.
n.next = n.next.next;
} else {
remove(index, current + 1, n.next);
}
}
public void remove(int index) {
if (first == null || index < 0) {
throw new IndexOutOfBoundsException();
}
if (index == 0) {
// Remove 'first'.
first = first.next;
} else {
remove(index, 0, first);
}
size--;
}
只需要一个索引:
private void remove(int index, Node n) {
if (n == null || index <= 0 || (index == 1 && n.next == null) {
throw new IndexOutOfBoundsException();
}
if (index == 1) {
// Remove 'n.next'.
n.next = n.next.next;
} else {
remove(index - 1, n.next);
}
}
public void remove(int index) {
if (first == null || index < 0) {
throw new IndexOutOfBoundsException();
}
if (index == 0) {
// Remove 'first'.
first = first.next;
} else {
remove(index, first);
}
size--;
}
Node
更好的是返回Node
而不是void
:
private Node remove(int index, Node n) {
if (n == null || index < 0) {
throw new IndexOutOfBoundsException();
}
if (index == 0) {
// Remove 'n' and return the rest of the list.
return n.next;
}
// 'n' stays. Update the rest of the list and return it.
n.next = remove(index - 1, n.next);
return n;
}
public void remove(int index) {
first = remove(index, first);
size--;
}