手记

玩转数据结构之链表

侠客岛的含笑
2018-07-15 08:48:13浏览 3511

之前介绍的动态数组，栈，队列底层都是依托于静态数组实现的，靠resize来解决固定容量问题。但是这次所说的链表是真正的动态数据结构（最简单的）

What‘s is Linked list?

Base Linked list

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();
        }
    }

}

Add Elements in linkedlist

prev在head初始化，然后寻找node之前的元素并标记

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);
    }
}

为链表设立虚拟头节点

去除Head添加元素的特殊性的方法就是设立虚拟头节点

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);
    }
}

链表遍历，查询和修改

  // 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();
    }
}
Test
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);
    }
}

删除元素

删除元素和添加元素基本类似，都是定位之前一个删除元素之前的元素
修改元素指向才是改变链表的本质（修改元素并不能改变链表）

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 --;
        }
    }

使用链表实现栈,有优势

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);
    }
}

Test(arrayStack linkedlistStack)

差异不是很大

队列

就像循环队列一样优化，我们使用带有尾指针的链表来改进链表
链表尾部删除元素还是需要O（n）的复杂度

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);
            }
        }
    }
}

链表与递归 Leetcode

链表天然的就带有递归的性质

/// Leetcode 203. Remove Linked List Elements
/// https://leetcode.com/problems/remove-linked-list-elements/description/

class Solution {

    public ListNode removeElements(ListNode head, int val) {

        while(head != null && head.val == val){
            ListNode delNode = head;
            head = head.next;
            delNode.next = null;
        }

        if(head == null)
            return head;

        ListNode prev = head;
        while(prev.next != null){
            if(prev.next.val == val) {
                ListNode delNode = prev.next;
                prev.next = delNode.next;
                delNode.next = null;
            }
            else
                prev = prev.next;
        }

        return head;
    }
}
利用虚拟头节点
/// Leetcode 203. Remove Linked List Elements
/// https://leetcode.com/problems/remove-linked-list-elements/description/

class Solution {

    public ListNode removeElements(ListNode head, int val) {

        while(head != null && head.val == val){
            ListNode delNode = head;
            head = head.next;
            delNode.next = null;
        }

        if(head == null)
            return head;

        ListNode prev = head;
        while(prev.next != null){
            if(prev.next.val == val) {
                ListNode delNode = prev.next;
                prev.next = delNode.next;
                delNode.next = null;
            }
            else
                prev = prev.next;
        }

        return head;
    }
}

Test

//Definition for singly-linked list.
public class ListNode {

    public int val;
    public ListNode next;

    public ListNode(int x) {
        val = x;
    }

    // 链表节点的构造函数
    // 使用arr为参数，创建一个链表，当前的ListNode为链表头结点
    public ListNode(int[] arr){

        if(arr == null || arr.length == 0)
            throw new IllegalArgumentException("arr can not be empty");

        this.val = arr[0];
        ListNode cur = this;
        for(int i = 1 ; i < arr.length ; i ++){
            cur.next = new ListNode(arr[i]);
            cur = cur.next;
        }
    }

    // 以当前节点为头结点的链表信息字符串
    @Override
    public String toString(){

        StringBuilder s = new StringBuilder();
        ListNode cur = this;
        while(cur != null){
            s.append(cur.val + "->");
            cur = cur.next;
        }
        s.append("NULL");
        return s.toString();
    }
}
    public static void main(String[] args) {

        int[] nums = {1, 2, 6, 3, 4, 5, 6};
        ListNode head = new ListNode(nums);
        System.out.println(head);

        ListNode res = (new Solution()).removeElements(head, 6);
        System.out.println(res);
    }
}

递归




递归微观解读

递归代价：函数调用+系统栈空间

输出调试

public class Solution {

    public ListNode removeElements(ListNode head, int val, int depth) {

        String depthString = generateDepthString(depth);

        System.out.print(depthString);
        System.out.println("Call: remove " + val + " in " + head);

        if(head == null){
            System.out.print(depthString);
            System.out.println("Return: " + head);
            return head;
        }

        ListNode res = removeElements(head.next, val, depth + 1);
        System.out.print(depthString);
        System.out.println("After remove " + val + ": " + res);

        ListNode ret;
        if(head.val == val)
            ret = res;
        else{
            head.next = res;
            ret = head;
        }
        System.out.print(depthString);
        System.out.println("Return: " + ret);

        return ret;
    }

    private String generateDepthString(int depth){
        StringBuilder res = new StringBuilder();
        for(int i = 0 ; i < depth ; i ++)
            res.append("--");
        return res.toString();
    }

    public static void main(String[] args) {

        int[] nums = {1, 2, 6, 3, 4, 5, 6};
        ListNode head = new ListNode(nums);
        System.out.println(head);

        ListNode res = (new Solution()).removeElements(head, 6, 0);
        System.out.println(res);
    }

}

双链表

循环链表

数组链表（明确知道链表的数量，有优势）

相关标签
Java
算法
2人推荐
随时随地看视频
慕课网APP

热门评论

兄弟，做这个用心了啊，把老师的PPT动画演示都进行截图放上来了，不像我，就只是做个简单的手记总结，hhh，继续加油

查看全部评论

相关阅读

命令行的贪吃蛇--集合的运用
Spring 一二事（5） - 依赖注入
双向链表的原理
玩转数据结构之字典树（前缀树 Trie）
链表