玩转数据结构 第4章 最基础的动态数据结构:链表

4-1 什么是链表

线性数据结构

为什么链表很重要?

  • 最简单的动态数据结构
  • 帮助更深入的理解引用(或者指针)
  • 更深入的理解递归
  • 辅助组成其他数据结构

链表 Linked List

数组和链表的对比

构造函数的写法

4-2 在链表中添加元素

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(E e){
            this(e, null);
        }

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

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

    private Node head;
    private int size;

    public LinkedList(){
        head = null;
        size = 0;
    }

    // 获取链表中的元素个数
    public int getSize(){
        return size;
    }

    // 返回链表是否为空
    public boolean isEmpty(){
        return size == 0;
    }

    // 在链表头添加新的元素e
    public void addFirst(E e){
//        Node node = new Node(e);
//        node.next = head;
//        head = node;

        head = new Node(e, head);
        size ++;
    }

    // 在链表的index(0-based)位置添加新的元素e
    // 在链表中不是一个常用的操作,练习用:)
    public void add(int index, E e){

        if(index < 0 || index > size)
            throw new IllegalArgumentException("Add failed. Illegal index.");

        if(index == 0)
            addFirst(e);
        else{
            Node prev = head;
            for(int i = 0 ; i < index - 1 ; i ++)
                prev = prev.next;

//            Node node = new Node(e);
//            node.next = prev.next;
//            prev.next = node;

            prev.next = new Node(e, prev.next);
            size ++;
        }
    }

    // 在链表末尾添加新的元素e
    public void addLast(E e){
        add(size, e);
    }
}
View Code

在链表头添加元素

在链表中间添加元素

4-3 使用链表的虚拟头结点 

 

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(E e){
            this(e, null);
        }

        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;
    }

    // 获取链表中的元素个数
    public int getSize(){
        return size;
    }

    // 返回链表是否为空
    public boolean isEmpty(){
        return size == 0;
    }

    // 在链表的index(0-based)位置添加新的元素e
    // 在链表中不是一个常用的操作,练习用:)
    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 ++;
    }

    // 在链表头添加新的元素e
    public void addFirst(E e){
        add(0, e);
    }

    // 在链表末尾添加新的元素e
    public void addLast(E e){
        add(size, e);
    }
}
03-DummyHead-in-LinkedList

4-4 链表的遍历,查询和修改 

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(E e){
            this(e, null);
        }

        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;
    }

    // 获取链表中的元素个数
    public int getSize(){
        return size;
    }

    // 返回链表是否为空
    public boolean isEmpty(){
        return size == 0;
    }

    // 在链表的index(0-based)位置添加新的元素e
    // 在链表中不是一个常用的操作,练习用:)
    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 ++;
    }

    // 在链表头添加新的元素e
    public void addFirst(E e){
        add(0, e);
    }

    // 在链表末尾添加新的元素e
    public void addLast(E e){
        add(size, e);
    }

    // 获得链表的第index(0-based)个位置的元素
    // 在链表中不是一个常用的操作,练习用:)
    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;
    }

    // 获得链表的第一个元素
    public E getFirst(){
        return get(0);
    }

    // 获得链表的最后一个元素
    public E getLast(){
        return get(size - 1);
    }

    // 修改链表的第index(0-based)个位置的元素为e
    // 在链表中不是一个常用的操作,练习用:)
    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;
    }

    // 查找链表中是否有元素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;
    }

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

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

        return res.toString();
    }
}
LinkedList

04-Query-and-Update-in-LinkedList

public class Main {

    public static void main(String[] args) {

        LinkedList<Integer> linkedList = new LinkedList<>();
        for(int i = 0 ; i < 5 ; i ++){
            linkedList.addFirst(i);
            System.out.println(linkedList);
        }

        linkedList.add(2, 666);
        System.out.println(linkedList);
    }
}
Main

4-5 从链表中删除元素 

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(E e){
            this(e, null);
        }

        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;
    }

    // 获取链表中的元素个数
    public int getSize(){
        return size;
    }

    // 返回链表是否为空
    public boolean isEmpty(){
        return size == 0;
    }

    // 在链表的index(0-based)位置添加新的元素e
    // 在链表中不是一个常用的操作,练习用:)
    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 ++;
    }

    // 在链表头添加新的元素e
    public void addFirst(E e){
        add(0, e);
    }

    // 在链表末尾添加新的元素e
    public void addLast(E e){
        add(size, e);
    }

    // 获得链表的第index(0-based)个位置的元素
    // 在链表中不是一个常用的操作,练习用:)
    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;
    }

    // 获得链表的第一个元素
    public E getFirst(){
        return get(0);
    }

    // 获得链表的最后一个元素
    public E getLast(){
        return get(size - 1);
    }

    // 修改链表的第index(0-based)个位置的元素为e
    // 在链表中不是一个常用的操作,练习用:)
    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;
    }

    // 查找链表中是否有元素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;
    }

    // 从链表中删除index(0-based)位置的元素, 返回删除的元素
    // 在链表中不是一个常用的操作,练习用:)
    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;
    }

    // 从链表中删除第一个元素, 返回删除的元素
    public E removeFirst(){
        return remove(0);
    }

    // 从链表中删除最后一个元素, 返回删除的元素
    public E removeLast(){
        return remove(size - 1);
    }

    // 从链表中删除元素e
    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();
    }
}
View Code

05-Remove-Element-in-LinkedList

public class Main {

    public static void main(String[] args) {

        LinkedList<Integer> linkedList = new LinkedList<>();
        for(int i = 0 ; i < 5 ; i ++){
            linkedList.addFirst(i);
            System.out.println(linkedList);
        }

        linkedList.add(2, 666);
        System.out.println(linkedList);

        linkedList.remove(2);
        System.out.println(linkedList);

        linkedList.removeFirst();
        System.out.println(linkedList);

        linkedList.removeLast();
        System.out.println(linkedList);
    }
}
Main

4-6 使用链表实现栈 

public interface Stack<E> {

    int getSize();
    boolean isEmpty();
    void push(E e);
    E pop();
    E peek();
}
public interface Stack

底层封装LinkedList实现的Stack

public class LinkedListStack<E> implements Stack<E> {

    private LinkedList<E> list;

    public LinkedListStack(){
        list = new LinkedList<>();
    }

    @Override
    public int getSize(){
        return list.getSize();
    }

    @Override
    public boolean isEmpty(){
        return list.isEmpty();
    }

    @Override
    public void push(E e){
        list.addFirst(e);
    }

    @Override
    public E pop(){
        return list.removeFirst();
    }

    @Override
    public E peek(){
        return list.getFirst();
    }

    @Override
    public String toString(){
        StringBuilder res = new StringBuilder();
        res.append("Stack: top ");
        res.append(list);
        return res.toString();
    }

    public static void main(String[] args) {

        LinkedListStack<Integer> stack = new LinkedListStack<>();

        for(int i = 0 ; i < 5 ; i ++){
            stack.push(i);
            System.out.println(stack);
        }

        stack.pop();
        System.out.println(stack);
    }
}
LinkedListStack

性能测试对比

import java.util.Random;

public class Main {

    // 测试使用stack运行opCount个push和pop操作所需要的时间,单位:秒
    private static double testStack(Stack<Integer> stack, int opCount){

        long startTime = System.nanoTime();

        Random random = new Random();
        for(int i = 0 ; i < opCount ; i ++)
            stack.push(random.nextInt(Integer.MAX_VALUE));
        for(int i = 0 ; i < opCount ; i ++)
            stack.pop();

        long endTime = System.nanoTime();

        return (endTime - startTime) / 1000000000.0;
    }

    public static void main(String[] args) {

        int opCount = 100000;

        ArrayStack<Integer> arrayStack = new ArrayStack<>();
        double time1 = testStack(arrayStack, opCount);
        System.out.println("ArrayStack, time: " + time1 + " s");

        LinkedListStack<Integer> linkedListStack = new LinkedListStack<>();
        double time2 = testStack(linkedListStack, opCount);
        System.out.println("LinkedListStack, time: " + time2 + " s");

        // 其实这个时间比较很复杂,因为LinkedListStack中包含更多的new操作
    }
}
Main

结论:时间复杂度的量级差异不太大。

4-7 带有尾指针的链表:使用链表实现队列 

public interface Queue<E> {

    int getSize();
    boolean isEmpty();
    void enqueue(E e);
    E dequeue();
    E getFront();
}
public interface Queue {
LinkedListQueue
public class LinkedListQueue<E> implements Queue<E> {

    private class Node{
        public E e;
        public Node 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 head, tail;
    private int size;

    public LinkedListQueue(){
        head = null;
        tail = null;
        size = 0;
    }

    @Override
    public int getSize(){
        return size;
    }

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

    @Override
    public void enqueue(E e){
        if(tail == null){
            tail = new Node(e);
            head = tail;
        }
        else{
            tail.next = new Node(e);
            tail = tail.next;
        }
        size ++;
    }

    @Override
    public E dequeue(){
        if(isEmpty())
            throw new IllegalArgumentException("Cannot dequeue from an empty queue.");

        Node retNode = head;
        head = head.next;
        retNode.next = null;
        if(head == null)
            tail = null;
        size --;
        return retNode.e;
    }

    @Override
    public E getFront(){
        if(isEmpty())
            throw new IllegalArgumentException("Queue is empty.");
        return head.e;
    }

    @Override
    public String toString(){
        StringBuilder res = new StringBuilder();
        res.append("Queue: front ");

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

    public static void main(String[] args){

        LinkedListQueue<Integer> queue = new LinkedListQueue<>();
        for(int i = 0 ; i < 10 ; i ++){
            queue.enqueue(i);
            System.out.println(queue);

            if(i % 3 == 2){
                queue.dequeue();
                System.out.println(queue);
            }
        }
    }
}
View Code

性能测试对比

import java.util.Random;

public class Main {

    // 测试使用q运行opCount个enqueueu和dequeue操作所需要的时间,单位:秒
    private static double testQueue(Queue<Integer> q, int opCount){

        long startTime = System.nanoTime();

        Random random = new Random();
        for(int i = 0 ; i < opCount ; i ++)
            q.enqueue(random.nextInt(Integer.MAX_VALUE));
        for(int i = 0 ; i < opCount ; i ++)
            q.dequeue();

        long endTime = System.nanoTime();

        return (endTime - startTime) / 1000000000.0;
    }

    public static void main(String[] args) {

        int opCount = 100000;

        ArrayQueue<Integer> arrayQueue = new ArrayQueue<>();
        double time1 = testQueue(arrayQueue, opCount);
        System.out.println("ArrayQueue, time: " + time1 + " s");

        LoopQueue<Integer> loopQueue = new LoopQueue<>();
        double time2 = testQueue(loopQueue, opCount);
        System.out.println("LoopQueue, time: " + time2 + " s");

        LinkedListQueue<Integer> linkedListQueue = new LinkedListQueue<>();
        double time3 = testQueue(linkedListQueue, opCount);
        System.out.println("LinkedListQueue, time: " + time3 + " s");
    }
}
Main

 

posted @ 2020-01-23 17:01  Marlon康  阅读(270)  评论(0编辑  收藏  举报