Singly linked list to doubly linked list
Question
A friend of mine wants me to convert his code into a doubly linked list, though I'm not familiar with it at all. I've looked up doubly linked lists but I can't tell by his code what to do with it. I'm not a master programmer. Do you have any suggestions?
import java.util.Collection;
import java.util.List;
class SinglyLinkedList<E> implements List<E> {
private class SinglyLinkedListNode<T> {
T data;
SinglyLinkedListNode<T> next;
public SinglyLinkedListNode() {
this(null, null);
}
public SinglyLinkedListNode(T data) {
this(data, null);
}
public SinglyLinkedListNode(T d, SinglyLinkedListNode<T> n) {
data = d;
next = n;
}
public boolean equals(Object o) {
if (data != null && o != null) {
return data.equals(((SinglyLinkedListNode) o).data);
} else {
return (data == null && o == null);
}
}
}
private SinglyLinkedListNode<E> list, last;
private int size;
public SinglyLinkedList() {
clear();
}
public void clear() {
list = last = null;
size = 0;
}
public boolean contains(Object o) {
SinglyLinkedListNode<E> t = list;
while (t != null) {
if (t.data == null) {
if (o == null) {
return true;
}
} else if (t.data.equals(o)) {
return true;
}
t = t.next;
}
return false;
}
public boolean add(E e) {
SinglyLinkedListNode<E> n = new SinglyLinkedListNode<E>(e);
if (isEmpty()) {
list = last = n;
} else {
last = last.next = n;
}
size++;
return true;
}
public void add(int index, E e) {
int currSize = size();
if (index < 0 || index > currSize) {
throw new IndexOutOfBoundsException(
"Index: " + index + ", Size: " + size());
}
if (isEmpty()) // index must == 0
{
list = last = new SinglyLinkedListNode<E>(e);
} else {
if (index == 0) {
list = new SinglyLinkedListNode<E>(e, list);
} else {
SinglyLinkedListNode<E> n = list;
for (int i = 0; i < index - 1; i++) {
n = n.next;
}
n.next = new SinglyLinkedListNode<E>(e, n.next);
if (index == currSize) {
last = n.next;
}
}
}
size++;
}
public boolean equals(SinglyLinkedList<E> e) {
SinglyLinkedListNode<E> e1 = list, e2 = e.list;
try {
for (int i = 1; i <= size(); i++) {
if (!e1.equals(e2)) {
return false;
}
e1 = e1.next;
e2 = e2.next;
}
} catch (NullPointerException ex) {
return false;
}
return true;
}
public E get(int index) {
if (index < 0 || index >= size()) {
throw new IndexOutOfBoundsException(
"Index: " + index + ", Size: " + size());
}
SinglyLinkedListNode<E> n = list;
int i = 0;
for (; i < index; i++) {
n = n.next;
}
return n.data;
}
@SuppressWarnings("unchecked")
public int indexOf(Object o) {
SinglyLinkedListNode<E> n = list;
int i = 0;
while (n != null) {
if ((o == null
? (n.data == null)
: (((E) o).equals(n.data)))) {
return i;
}
n = n.next;
i++;
}
return -1;
}
public boolean isEmpty() {
return list == null;
}
public E remove(int index) {
if (index < 0 || index >= size()) {
throw new IndexOutOfBoundsException(
"Index: " + index + ", Size: " + size());
}
SinglyLinkedListNode<E> n = list, prevNode = null;
int i = 0;
while (true) {
if (index == i) {
if (n == list) // removing first node
{
list = list.next;
} else {
prevNode.next = n.next;
}
if (n == last) {
last = prevNode;
}
size--;
return n.data;
}
prevNode = n;
n = n.next;
i++;
}
}
@SuppressWarnings("unchecked")
public boolean remove(Object o) {
SinglyLinkedListNode<E> n = list, prevNode = null;
while (n != null) {
if ((o == null
? (n.data == null)
: (((E) o).equals(n.data)))) {
if (n == list) //removing first node
{
list = list.next;
} else {
prevNode.next = n.next;
}
if (n == last) {
last = prevNode;
}
size--;
return true;
}
prevNode = n;
n = n.next;
}
return false;
}
public int size() {
return size;
}
public String toString() {
String s = "((";
SinglyLinkedListNode<E> t = list;
if (t != null) {
while (t.next != null) {
s += t.data + ", ";
t = t.next;
}
s += last.data;
}
return s + "))";
}
5 answers
solution1
1 ACCPTED 2010-11-17 20:03:08
I don't understand the problem. If this is homework, you should say so -- community rules! A quick explanation, regardless:
A linked list is a structure with the following,... well, structure:
DATA |--> DATA
REFERENCE TO NEXT ITEM ---| REFERENCE TO NEXT ITEM ---...
Each "link" in the "chain" contains some data, and a way to locate the next item in the chain. That's a singly linked list, as you said.
A doubly linked list is a very similar structure, only each link in the chain contains both a way of locating the next item, and a way of locating the previous item. If you need to be able to walk the list both ways, you'll need this kind of structure.
|-> DATA |--> DATA
| REFERENCE TO NEXT ITEM ---| REFERENCE TO NEXT ITEM ---...
---------------------------------- REFERENCE TO PREV ITEM
Ooookay the "drawings" are hideous. You can look up what a doubly linked list is with a Google query and get better information, on second thought, but oh well.
solution2
0 2010-11-17 20:03:30
Take a look at this other question on SO. It should help you to better understand the doubly linked list.
solution3
0 2010-11-17 20:03:40
每个节点都需要一个next还有一个previous占位符，它是一个双向链表。
solution4
0 2010-11-17 20:03:58
LinkedList in Java is a doubly-linked list. Why would you want to create one by yourself?
solution5
0 2010-11-17 20:04:27
sounds like homework. This will get you started.
http://en.wikipedia.org/wiki/Doubly_linked_list
basically, in a singly linked list each node has a pointer to the next node. In a doubly linked list there are pointers to both next and previous. Beware how the pointers work at the beginning and the end of the list.
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