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Java에서 노드를 사용하여 이중 연결 목록을 구현했습니다. 노드 및 연결된 목록 클래스에 대한 내 코드는 아래와 같습니다. 작업의 디자인 및/또는 실행 시간을 개선하기위한 제안 사항은 무엇입니까?Java에서 이중 링크 목록 구현의 효율성
public class Node implements Comparable<Node>{
Node next;
Node previous;
Object element;
int index;
public Node(Object element){
this.element = element;
this.next = null;
this.previous = null;
}
public Node(Object element, Node next, Node previous){
this.element = element;
this.next = next;
this.previous = previous;
}
public Node(Object element, Node next, Node previous, int index){
this.element = element;
this.next = next;
this.previous = previous;
this.index = index;
}
//Getter Method
public Node get_next(){
return this.next;
}
//For comparable class
public int compareTo(Node x){
if (this.element.equals(x.element)){
return 0;
}
else
return 1;
}
public boolean compare_index(int index){
if (this.index == index){
return true;
}
else
return false;
}
}
public class Mylinkedlists{
Node first_node;
Node last_node;
int size = 0;
//Constructors
public Mylinkedlists(Object element){
Node temp = new Node(element, last_node,first_node,0);
first_node = new Node (null,temp,null);
last_node = new Node(null,null,temp);
size+=1;
}
public Mylinkedlists(){
first_node = new Node (null,last_node,null);
last_node = new Node(null,null,first_node);
}
//Returns size of Node
public int size(){
return size;
}
//Adds element to end of linked list
public void add(Object element){
Node to_add = new Node(element);
Node temp = last_node.previous;
size+=1;
//Pointer changes
last_node.previous.next = to_add;
last_node.previous = to_add;
to_add.previous = temp;
to_add.next = last_node;
to_add.index = size-1;
}
//Inserts Element after node
public void insert(Object add_after, Object element_to_add) throws NotFoundException{
Node current = first_node.next;
Node insert_after = new Node(add_after);
Node to_add = new Node(element_to_add);
//find the node
while (current.compareTo(insert_after) != 0){
current = current.next;
}
//Inserts element after node
if (current.compareTo(insert_after) == 0){
Node temp = current.next;
current.next.previous = to_add;
to_add.next = temp;
current.next = to_add;
to_add.previous = current;
size+=1;
}
else
throw new NotFoundException("Element not in list");
}
//Removes element from linked list
public void remove(Object element) throws NotFoundException{
boolean stop = false;
Node search_for = new Node(element);
Node current = first_node.next;
for (int i = 0; i < size && !stop; i ++){
//Compares nodes
if (current.compareTo(search_for) == 1){
current.next.previous = current.previous;
current.previous.next = current.next;
current.next = null;
current.previous = null;
size-=1;
stop = true;
}
else
current = current.next;
}
//If element not in list
if (stop == false && current.next == null)
throw new NotFoundException("Element not in list");
}
//Removes element from end of linked list
public void remove_From_End(){
last_node.previous.next = null;
Node temp = last_node.previous.previous;
last_node.previous.previous = null;
last_node.previous = temp;
last_node.previous.next = last_node;
size -= 1;
}
//Returns true if list contains element, else false
public boolean contains(Object element){
boolean contains = false;
Node current = first_node.next;
Node with_element = new Node(element);
while (current.next != null){
if (current.compareTo(with_element) == 0)
return true;
else
current = current.next;
}
return contains;
}
public Object get(int index) throws NotFoundException{
if (index < 0 || index > (size-1))
throw new NotFoundException("Index out of range");
Node current = first_node;
for (int i = 0; i <= index; i++){
current = current.next;
}
return current.element;
}
public String toString(){
String temp = "";
Node temp2 = first_node;
for (int i = 0 ; i < size; i++){
temp2 = temp2.next;
temp += String.valueOf(temp2.element) + " ";
}
return temp;
}
}
예를 들어, 도트 (.) 표기법을 사용하여 직접 Node 클래스의 요소에 액세스합니다. getter 및 setter 메서드가 유일한 대안입니까? 게다가, O (n) 시간보다 적게 걸리는 get 메소드의 다른 구현이 있습니까?
Node 클래스에서 수행 한 것처럼 점 표기법을 사용하는 것이 좋습니다. – Zaghie
점 표기법을 사용합니다. – Degustaf