Java 编译器不会在单链表的递归反转中前进
我在SingleLinkedList类中有一个递归静态方法,该类由于一个不确定的原因而永远运行。此类是泛型的,其声明如下所示:Java 编译器不会在单链表的递归反转中前进,java,recursion,singly-linked-list,Java,Recursion,Singly Linked List,我在SingleLinkedList类中有一个递归静态方法,该类由于一个不确定的原因而永远运行。此类是泛型的,其声明如下所示: public class SinglyLinkedList<E>{ private static class Node<E> { private E element; private Node<E> next; public Node(E e, Node<E> n) { this
public class SinglyLinkedList<E>{
private static class Node<E> {
private E element;
private Node<E> next;
public Node(E e, Node<E> n) {
this.element = e;
this.next = n;
}
public E getElement() {
return this.element;
}
public Node<E> getNext() {
return this.next;
}
public void setNext(Node<E> n) {
this.next = n;
}
}
private Node<E> removeFirstNode() {
if (isEmpty()) return null;
Node<E> answer = this.head;
this.head = this.head.getNext();
answer.next = null; // Set next ptr to null
this.size--;
if (this.size == 0) {
this.tail = null;
}
return answer;
}
此外,SinglyLinkedList类具有以下字段:
private Node<E> head = null;
private Node<E> tail = null;
private int size = 0;
反向方法使用名为removeFirstNode的非静态方法,其目的是删除单链表中的第一个节点并返回它:
private Node<E> removeFirstNode() {
if (isEmpty()) return null;
//Node<E> answer = new Node<>(this.head.getElement(), null);
Node<E> answer = this.head;
this.head = this.head.getNext();
this.size--;
if (this.size == 0) this.tail = null;
return answer;
}
问题是,当我尝试在大小等于2的SingleLinkedList上运行reverse方法时,编译器会到达该行
reversed.addLast(first);
reversed.addLast(first);
然后在addLast方法中,它停在行上
this.tail = n;
this.tail = n;
永远不停地奔跑。如果大小等于3或更多,编译器将到达该行
reversed.addLast(first);
reversed.addLast(first);
并在那里停止,甚至没有输入addLast方法。现在,如果我更换线路
Node<E> answer = this.head;
编辑2:
最小代码:
public class SinglyLinkedList<E>{
private static class Node<E> {
private E element;
private Node<E> next;
public Node(E e, Node<E> n) {
this.element = e;
this.next = n;
}
public E getElement() {
return this.element;
}
public Node<E> getNext() {
return this.next;
}
public void setNext(Node<E> n) {
this.next = n;
}
}
private Node<E> head = null;
private Node<E> tail = null;
private int size = 0;
public SinglyLinkedList() {}
public int size() { return this.size; }
public boolean isEmpty() {
return this.size == 0;
}
public void addLast(E e) {
Node<E> newest = new Node<>(e, null);
if (isEmpty())
this.head = newest;
else
this.tail.setNext(newest);
this.tail = newest;
this.size++;
}
@Override
public String toString() {
String output = "";
if (this.size > 0) {
Node<E> current_node = head;
while (current_node != null) {
output += current_node.getElement();
if (current_node != tail) output += ", ";
else output += "\n";
current_node = current_node.getNext();
}
}
return output;
}
private void addLast(Node<E> n) {
if (isEmpty()) this.head = n;
else this.tail.setNext(n);
this.tail = n;
this.size++;
}
private Node<E> removeFirstNode() {
if (isEmpty()) return null;
//Node<E> answer = new Node<>(this.head.getElement(), null);
Node<E> answer = this.head;
this.head = this.head.getNext();
this.size--;
if (this.size == 0) this.tail = null;
return answer;
}
public static SinglyLinkedList reverse(SinglyLinkedList l) {
if (l.size < 2) return l;
Node first = l.removeFirstNode();
SinglyLinkedList reversed = reverse(l);
reversed.addLast(first);
return reversed;
}}
测试等级:
public static void main(String[] args) {
int n = 4;
SinglyLinkedList<Integer> list = new SinglyLinkedList<>();
for (int i = 1; i <= n; i++) list.addLast(i);
System.out.print(list);
System.out.print(SinglyLinkedList.reverse(list));
}
removeFirstNode的实现不会取消该节点的下一个指针的链接
在原始链表中,第一个节点的下一个点指向第二个节点,第二个节点的下一个指针为null
在新列表中,第一个节点的下一个指针将指向第二个节点,但第二个节点的下一个指针将指向以前是第二个节点的第一个节点
类似于原始列表:
+---+ +---+
| A |--->| B |--->null
+---+ +---+
当重新排序时,由于A的下一个指针仍然指向B,因此变为:
+---+ +---+
| B |--->| A |---+
+---+ +---+ |
^ |
| |
+--------------+
您可以将removeFirstNode实现更改为以下内容:
public class SinglyLinkedList<E>{
private static class Node<E> {
private E element;
private Node<E> next;
public Node(E e, Node<E> n) {
this.element = e;
this.next = n;
}
public E getElement() {
return this.element;
}
public Node<E> getNext() {
return this.next;
}
public void setNext(Node<E> n) {
this.next = n;
}
}
private Node<E> removeFirstNode() {
if (isEmpty()) return null;
Node<E> answer = this.head;
this.head = this.head.getNext();
answer.next = null; // Set next ptr to null
this.size--;
if (this.size == 0) {
this.tail = null;
}
return answer;
}
代码看起来可能会停止,因为调试器尝试使用toString打印列表,我想它会遍历列表,并且由于循环而从未完成。请提供可以生成和调试的完整代码。看看这个主意。此外,您可能希望熟悉调试器,这种短代码可以逐行运行,然后您将看到发生了什么以及如何进行。您如何知道JVM在某一行上“莫名其妙地终止”?你是在使用调试器还是什么?是的,我正在使用调试器。我相信这就是问题所在,因此我提出了注释掉的解决方案,但我仍然想知道:这不是循环链接列表背后的概念吗?如果是这样,为什么会导致问题?啊,toString方法中的while循环导致了问题。就这样!