为什么我的删除方法会从我的双向链接列表中删除每个元素？如果我取出if / else语句，那么我可以成功删除中间元素，但列表的头部或尾部的元素仍然存在。但是，我添加了if / else语句来处理头部和尾部的元素，不幸的是，此方法现在删除了列表中的每个元素。我做错了什么？
package week6;
import java.util.Iterator;
public class DblLinkedList<E>
{
private LinkEntry<E> head = null;
private LinkEntry<E> tail = null;
private int size = 0;
public DblLinkedList()
{
head = tail = null;
}
public boolean is_empty()
{
if (head == null)
return true;
return false;
}
public int size()
{
int count = 0;
for (LinkEntry<E> current = head; current != null; current = current.next)
count++;
return count;
}
public boolean add(E e)
{
LinkEntry<E> new_element = new LinkEntry<E>();
new_element.element = e;
if (head == null)
{
new_element.next = head;
head = new_element;
tail = head;
}
else
{
tail.next = new_element;
new_element.previous = tail;
tail = new_element;
}
return true;
}
public void remove(int n)
{
LinkEntry<E> remove_this = new LinkEntry<E>();
//if nothing comes before remove_this, set the head to equal the element after remove_this
if (remove_this.previous == null)
head = remove_this.next;
//if nothing comes after remove_this, set the tail equal to the element before remove_this
else if (remove_this.next == null)
tail = remove_this.previous;
//otherwise set the next element's previous pointer to the element before remove_this
else
{
//if remove_this is located in the middle of the list, enter this loop until it is
//found, then remove it, closing the gap afterwards.
int i = 0;
for (remove_this = head; remove_this != null; remove_this = remove_this.next)
{
//if i == n, stop and delete 'remove_this' from the list
if (i == n)
{
//set the previous element's next to the element that comes after remove_this
remove_this.previous.next = remove_this.next;
//set the element after remove_this' previous pointer to the element before remove_this
remove_this.next.previous = remove_this.previous;
break;
}
//if i != n, keep iterating through the list
i++;
}
}
}
/*
* Print the doubly linked list starting at the beginning.
*/
public void print_from_beginning()
{
LinkEntry<E> current = new LinkEntry<E>();
for (current = head; current != null; current = current.next)
{
System.out.print(current.element + " ");
}
}
/*
* Print the doubly linked list starting the end.
*/
public void print_from_end()
{
LinkEntry<E> current = new LinkEntry<E>();
for (current = tail; current != null; current = current.previous)
{
System.out.print(current.element + " ");
}
}
/* ------------------------------------------------------------------- */
/* Inner classes */
protected class LinkEntry<E>
{
protected E element;
protected LinkEntry<E> next;
protected LinkEntry<E> previous;
protected LinkEntry() { element = null; next = previous = null; }
}
/* ------------------------------------------------------------------- */
protected class DblLinkedListImplIterate<E> implements Iterator<E>
{
protected LinkEntry<E> next;
protected DblLinkedListImplIterate()
{
next = (LinkEntry<E>) head;
}
@Override
public boolean hasNext() {
// TODO Auto-generated method stub
return false;
}
@Override
public E next() {
// TODO Auto-generated method stub
return null;
}
@Override
public void remove() {
// TODO Auto-generated method stub
}
}
}
我测试我的方法的主要课程是：
package week6;
public class App {
public static <E> void main(String[] args) {
DblLinkedList<String> list = new DblLinkedList<String>();
list.add("Bill");
list.add("Rohan");
list.add("James");
list.add("Krishna");
list.add("Javier");
list.add("Lisa");
System.out.println("List size after all names are added: " + list.size());
//a. Print the linked list starting at the beginning.
System.out.println("\nPrint the linked list starting at the beginning:");
list.print_from_beginning();
System.out.println();
//b. Print the linked list starting at the end.
System.out.println("\nPrint the linked list starting at the end:");
list.print_from_end();
System.out.println();
//c. Remove Bill and print the linked list starting from beginning.
System.out.println("\nRemove Bill and print the linked list starting from beginning:");
list.remove(1);
list.print_from_beginning();
System.out.println();
//d. Remove Lisa and print the linked list starting from end.
System.out.println("\nRemove Lisa and print the linked list starting from end:");
list.remove(5);
list.print_from_end();
System.out.println();
//e. Remove Krishna and print the linked list starting from the beginning.
System.out.println("\nRemove Krishna and print the linked list starting from the beginning:");
list.remove(2);
list.print_from_beginning();
System.out.println();
System.out.println("\nList size: " + list.size());
}
}
运行程序后输出的结果是：
List size after all names are added: 6
Print the linked list starting at the beginning:
Bill Rohan James Krishna Javier Lisa
Print the linked list starting at the end:
Lisa Javier Krishna James Rohan Bill
Remove Bill and print the linked list starting from beginning:
Remove Lisa and print the linked list starting from end:
Lisa Javier Krishna James Rohan Bill
Remove Krishna and print the linked list starting from the beginning:
List size: 0
答案 0 :(得分：1)
通过添加两个if块来处理第一个和最后一个，除了以前之外，还添加了2个可执行块。现在所有三个区块都在一起执行。我相信你正在尝试做这样的事情（一次只执行一个块）：
public void remove(int n)
{
LinkEntry<E> remove_this = new LinkEntry<E>();
//if remove_this is located in the middle of the list, enter this loop until it is
//found, then remove it, closing the gap afterwards.
int i = 0;
boolean removed = false;
remove_this = head;
while(removed == false){
//if nothing comes before remove_this, set the head to equal the element after remove_this
if (remove_this.previous == null){
head = remove_this.next;
head.previous = null;
removed = true;
}
//if nothing comes after remove_this, set the tail equal to the element before remove_this
else if (remove_this.next == null){
tail = remove_this.previous;
tail.next = null;
removed = true;
}
//otherwise set the next element's previous pointer to the element before remove_this
else{
//if i == n, stop and delete 'remove_this' from the list
if (i == n) {
//set the previous element's next to the element that comes after remove_this
remove_this.previous.next = remove_this.next;
//set the element after remove_this' previous pointer to the element before remove_this
remove_this.next.previous = remove_this.previous;
removed = true;
break;
}
//if i != n, keep iterating through the list
}
if(!removed){
remove_this = remove_this.next;
}
i++;
}
}
请注意：当您在前head个条件中分配tail或if else时，remove_this为空，因此会head或tail为null并且您打印的程序没有打印任何内容。