[英]Generic linked-list remove, size, get methods
我剛剛在 Stackoverflow 上看到了這個問題“Java 中的通用鏈接列表”中的精彩代碼。 我一直在思考如何實現方法 remove(從鏈表中刪除單個節點)、size(獲取列表的大小)和 get(獲取 a 節點)。 有人可以告訴我怎么做嗎?
public class LinkedList<E> {
private Node head = null;
private class Node {
E value;
Node next;
// Node constructor links the node as a new head
Node(E value) {
this.value = value;
this.next = head;//Getting error here
head = this;//Getting error here
}
}
public void add(E e) {
new Node(e);
}
public void dump() {
for (Node n = head; n != null; n = n.next)
System.out.print(n.value + " ");
}
public static void main(String[] args) {
LinkedList<String> list = new LinkedList<String>();
list.add("world");
list.add("Hello");
list.dump();
}
}
您對操作remove()
、 size()
和contains()
的 LinkedList 實現如下所示:
static class LinkedList<Value extends Comparable> {
private Node head = null;
private int size;
private class Node {
Value val;
Node next;
Node(Value val) {
this.val = val;
}
}
public void add(Value val) {
Node oldHead = head;
head = new Node(val);
head.next = oldHead;
size++;
}
public void dump() {
for (Node n = head; n != null; n = n.next)
System.out.print(n.val + " ");
System.out.println();
}
public int size() {
return size;
}
public boolean contains(Value val) {
for (Node n = head; n != null; n = n.next)
if (n.val.compareTo(val) == 0)
return true;
return false;
}
public void remove(Value val) {
if (head == null) return;
if (head.val.compareTo(val) == 0) {
head = head.next;
size--;
return;
}
Node current = head;
Node prev = head;
while (current != null) {
if (current.val.compareTo(val) == 0) {
prev.next = current.next;
size--;
break;
}
prev = current;
current = current.next;
}
}
}
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