带虚拟头结点的链表的实现

带虚拟头结点的链表的实现

  1. 链表是一种动态的线性数据结构。数据存储在节点内。
  2. 链表与数组的对比
  • 数组最好用于索引有语意的情况
  • 数组的最大优点在于支持随机访问
  • 链表不适合用于索引有语义的情况
  • 链表最大的特点是动态,但是不支持随机访问
  1. 带虚拟头结点链表的实现
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package LinkedList;

/**
* @ Description: 带虚拟节点的链表
* @ Date: Created in 13:23 12/07/2018
* @ Author: Anthony_Duan
*/
public class LinkedList<E> {

private class Node {

public E e;

public Node next;

public Node(E e, Node next) {
this.e = e;
this.next = next;
}

public Node() {
this(null, null);
}


@Override
public String toString() {
return e.toString();
}
}


private Node dummyHead;
private int size;

public LinkedList() {
dummyHead = new Node();
size = 0;
}

public int getSize() {
return size;
}

public boolean isEmpty() {
return size == 0;
}

public void add(int index, E e) {
if (index < 0 || index > size) {
throw new IllegalArgumentException("Add failed. Illegal index.");
}
Node prev = dummyHead;
for (int i = 0; i < index; i++) {
prev = prev.next;
}
prev.next = new Node(e, prev.next);
size++;
}

public void addFirst(E e) {
add(0, e);
}

public void addLast(E e) {
add(size, e);
}


public E get(int index) {
if (index < 0 || index >= size) {
throw new IllegalArgumentException("Get failed. Illegal index.");
}
Node cur = dummyHead.next;
for (int i = 0; i < index; i++) {
cur = cur.next;
}
return cur.e;
}

public E getFirst() {
return get(0);
}

public E getLast() {
return get(size);
}

public void set(int index, E e) {
if (index < 0 || index >= size) {
throw new IllegalArgumentException("Set failed. Illegal index.");
}
Node cur = dummyHead.next;
for (int i = 0; i < index; i++) {
cur = cur.next;
}
cur.e = e;
}


public boolean contains(E e) {
Node cur = dummyHead.next;
while (cur != null) {
if (cur.e.equals(e)) {
return true;
}
cur = cur.next;
}
return false;
}

public E remove(int index) {
if (index < 0 || index >= size) {
throw new IllegalArgumentException("Remove failed. Index is illegal.");
}
Node prev = dummyHead;
for (int i = 0; i < index; i++) {
prev = prev.next;
}

Node retNode = prev.next;
prev.next = retNode.next;
retNode.next = null;
size--;
return retNode.e;
}


public E removeFirst() {
return remove(0);
}

public E removeLast() {
return remove(size - 1);
}

public void removeElement(E e) {
Node prev = dummyHead;
while (prev.next != null) {
if (prev.next.e.equals(e)) {
break;
}
prev = prev.next;
}

if (prev.next != null) {
Node delNode = prev.next;
prev.next = delNode.next;
delNode.next = null;
size--;
}


}

@Override
public String toString() {
StringBuilder res = new StringBuilder();

Node cur = dummyHead.next;
while (cur != null) {
res.append(cur + "->");
cur = cur.next;
}
res.append("NULL");

return res.toString();
}
}
  1. 链表的时间复杂度分析

添加操作

  • addLast(e);——————————O(n)
  • addFirst(e);——————————O(1)
  • add(index,e);————————O(n/2)=O(n)

删除操作

  • removeLast(e)————————O(n)
  • removeFirst(e)————————O(1)
  • remove(index,e)————————O(n/2)=O(n)

修改操作

  • set(index,e)————————O(n)

查找操作

  • get(index)——————————O(n)
  • contains(e)——————————O(n)
    总的来说对链表表头的操作的时间复杂度为O(1)
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