資料結構 - 基於連結串列的佇列
基於連結串列的佇列
當我們基於連結串列實現佇列時,需要從一端加元素,另一端取出元素,就需要引入一個新的變數tail指向連結串列的尾部,此時,向尾部進行新增操作時間複雜度會變為O(1),然而刪除操作還是需要從head向後遍歷,所以此時選擇連結串列尾為隊尾,連結串列頭為隊首。
基於連結串列的實現的原始碼如下:
package queue;
import linkedList.LinkedList;
public class LinkedListQueue<E> implements Queue<E> {
private class Node<E>{
E e;
Node<E> next;
public Node(E e,Node next){
this.e = e;
this.next = next;
}
public Node(E e){
this(e, null);
}
public Node(){
this(null, null);
}
@Override
public String toString() {
return e.toString();
}
}
private Node<E> dummyHead; //虛擬頭結點
private Node<E> tail; //指向連結串列尾節點的索引
private int size; //連結串列中元素的個數
public LinkedListQueue(){
dummyHead = new Node<>();
tail = dummyHead;
size = 0;
}
@Override
public int getSize() {
return size;
}
@Override
public boolean isEmpty() {
return size == 0;
}
@Override
public void enqueue(E e) {
tail.next = new Node<E>(e);
tail = tail.next;
size++;
}
@Override
public E dequeue() {
if (isEmpty()) {
throw new IllegalArgumentException("Dequeue failed.Dequeue is empty.");
}
Node<E> retNode = dummyHead.next;
dummyHead.next = retNode.next;
retNode.next = null;
if (dummyHead.next == null) {
tail = dummyHead;
}
size--;
return retNode.e;
}
@Override
public E getFront() {
if (isEmpty()) {
throw new IllegalArgumentException("GetFront failed.Queue is Empty.");
}
return dummyHead.next.e;
}
@Override
public String toString() {
StringBuilder res = new StringBuilder();
Node<E> cur = dummyHead.next;
res.append("Queue: front ");
while (cur != null) {
res.append(cur + "->");
cur = cur.next;
}
res.append("NULL ");
res.append("tail");
return res.toString();
}
}
基於連結串列佇列的時間複雜度的分析:
*void enqueue(E e) : O(1)
*E dequeue() : O(1)
*E getFront() : O(1)
*E getSize() : O(1)
*boolean isEmpty() : O(1)