一、特点

  1. 链表是以节点的方式来存储，是链式存储。
  2. 每个节点包含data域和next域，next域指向下一个节点。
  3. 链表的各个节点不一定是连续存放的。
  4. 链表分为带头结点的链表和没有头结点的链表，根据实际需求确定。

内存结构图
image.png
逻辑结构图
image.png

二、代码实现

package a03.linkedList;

public class SingleLinkedList {

    //先初始化一个头结点，头结点不动，不存放具体的数据，用来表示单链表头
    private HeroNode head = new HeroNode(0, "", "");

    //添加节点，直接加到末尾
    public void add(HeroNode heroNode) {
        //因为head节点不能动，需要一个辅助节点
        HeroNode temp = head;
        //遍历链表找到最后
        while (true) {
            if (temp.next == null) {
                break;
            }
            temp = temp.next;
        }
        //将节点接入链表
        temp.next = heroNode;
    }

    //添加节点，按编号顺序添加
    public void addByOrder(HeroNode heroNode) {
        HeroNode temp = head;

        boolean flag = false; //标志添加的编号是否存在
        while (true) {
            if (temp.next == null) {
                break;
            }
            //因为是单链表，找的temp是位于添加位置的前一个节点。
            if (temp.next.no > heroNode.no) {
                break;
            } else if (temp.next.no == heroNode.no) {
                flag = true;
                break;
            } else {
                temp = temp.next;
            }
        }
        if (flag) {
            System.out.println("不能添加，编号已存在: " + heroNode);
        } else {
            heroNode.next = temp.next;
            temp.next = heroNode;
        }
    }

    //修改链表
    public void update(HeroNode hero) {
        if (head.next == null) {
            System.out.println("链表为空");
            return;
        }
        HeroNode temp = head;
        //标志有没找到对应的节点
        boolean flag = false;
        while (true) {
            if (temp == null) {
                break;
            }
            if (temp.no == hero.no) {
                flag = true;
                break;
            }
            temp = temp.next;
        }
        if (flag) {
            temp.name = hero.name;
            temp.nickName = hero.nickName;
        } else {
            System.out.println("没有找到该编号的节点：" + hero);
        }
    }


    //删除节点
    public void delete(int no){
        HeroNode temp = head;

        boolean flag = false; //标志要删除的编号是否存在
        while (true) {
            if (temp.next == null) {
                break;
            }
            //因为是单链表，找的temp是位于添加位置的前一个节点。
            if (temp.next.no == no) {
                flag = true;
                break;
            } else if (temp.next.no > no) {
                break;
            } else {
                temp = temp.next;
            }
        }
        if (!flag) {
            System.out.println("不能添加，编号不存在: " + no);
        } else {
            temp.next = temp.next.next;
        }
    }

    //显示链表
    public void list() {
        if (head.next == null) {
            System.out.println("链表为空");
            return;
        }
        HeroNode temp = head.next;
        while (true) {
            if (temp == null) {
                break;
            } else {
                System.out.println(temp);
                temp = temp.next;
            }
        }
    }

    public static void main(String[] args) {
        HeroNode hero1 = new HeroNode(1, "宋江", "及时雨");
        HeroNode hero2 = new HeroNode(2, "卢俊义", "玉麒麟");
        HeroNode hero3 = new HeroNode(3, "吴用", "智多星");
        HeroNode hero4 = new HeroNode(4, "林冲", "豹子头");
        HeroNode updateHero = new HeroNode(2, "小卢", "玉麒麟");

        SingleLinkedList singleLinkedList = new SingleLinkedList();
        singleLinkedList.addByOrder(hero3);
        singleLinkedList.addByOrder(hero2);
        singleLinkedList.addByOrder(hero4);
        singleLinkedList.addByOrder(hero1);
        singleLinkedList.addByOrder(hero1);
        singleLinkedList.update(updateHero);
        singleLinkedList.delete(2);
        singleLinkedList.list();

    }
}

class HeroNode {
    int no;
    String name;
    String nickName;
    HeroNode next;

    public HeroNode(int no, String name, String nickName) {
        this.no = no;
        this.name = name;
        this.nickName = nickName;
    }

    @Override
    public String toString() {
        return "HeroNode{" +
                "no=" + no +
                ", name='" + name + '\'' +
                ", nickName='" + nickName + '\'' +
                '}';
    }
}

三、练习题

3.1 求单链表中节点的个数

//获取单链表有效节点个数
    public int size(HeroNode head){
        //如果带头节点的链表，不统计头结点
        if(head.next == null){
            return 0;
        }
        int count = 0;
        HeroNode temp = head.next;
        while(true){
            if(temp == null){
                break;
            }
            count++;
            temp = temp.next;
        }
        return count;
    }

3.2 求单链表中倒数第k个节点

//查找单链表中倒数第n个节点
    public HeroNode reciprocal(HeroNode head, int index) {
        //获取单链表有效节点个数
        int size = size(head);
        if(index <=0 || index >size){
            return null;
        }
        HeroNode temp = head.next;
        for(int i=0; i<size-index; i++){
            temp = temp.next;
        }
        return temp;
    }

3.3 单链表的反转

    //反转链表
    public void reverse(HeroNode head){
        if(head.next == null || head.next.next ==null){
            return ;
        }
        //帮助遍历原链表
        HeroNode temp = head.next;
        HeroNode next = null;
	//另建一个链表头来保存结果
        HeroNode reverseHead =  new HeroNode(0,"","");
        while(true){
            if(temp == null){
                break;
            }
            next = temp.next;
            temp.next = reverseHead.next;
            reverseHead.next = temp;
            temp = next;
        }
        head.next = reverseHead.next;
    }

3.4 从尾到头打印单链表

    //从尾到头打印链表
    public void showReverse(HeroNode head){
        if(head == null){
            return;
        }
        HeroNode temp = head.next;
        Stack<HeroNode> stock = new Stack<>();
        while(true){
            if(temp == null){
                break;
            }
            stock.push(temp);
            temp = temp.next;
        }
        while (!stock.isEmpty()){
            System.out.println(stock.pop());
        }
    }

3.5 合并两个有序的单链表

    //合并两个有序的单向链表
    public HeroNode merge(HeroNode head1, HeroNode head2){
        if(head1 == null){
            return head2;
        }
        if(head2 == null){
            return head1;
        }

        HeroNode result = new HeroNode(0,"","");
        HeroNode temp1 = head1.next;
        HeroNode temp2 = head2.next;

        while(true){
            if(temp1 == null || head2 == null) {
                break;
            }else if(temp1.no <= head2.no){
                result.next=temp1;
                temp1 = temp1.next;
            }else{
                result.next=head2;
                temp2 = temp2.next;
            }
        }
        while(true){
            if(temp1 == null){
                break;
            }
            result.next=temp1;
            temp1 = temp1.next;
        }
        while(true){
            if(temp2 == null){
                break;
            }
            result.next=temp2;
            temp2 = temp2.next;
        }
        return result.next;
    }

四、双向链表

4.1 单链表的不足

  1. 查找的方向只能是一个，而双向链表可以向前或者向后查找。
  2. 单向链表不能自我删除，需要靠辅助节点，而双向链表可以自我删除。

4.2 代码实现

package a03.linkedList;

public class DoubleLinkedList {
    //头结点
    private PersonNode head = new PersonNode(0, "", "");

    public PersonNode getHead() {
        return head;
    }

    //显示链表
    public void list() {
        if (head.next == null) {
            System.out.println("链表为空");
            return;
        }
        PersonNode temp = head.next;
        while (true) {
            if (temp == null) {
                break;
            } else {
                System.out.println(temp);
                temp = temp.next;
            }
        }
    }

    //添加节点，直接加到末尾
    public void add(PersonNode person) {
        //因为head节点不能动，需要一个辅助节点
        PersonNode temp = head;
        //遍历链表找到最后
        while (true) {
            if (temp.next == null) {
                break;
            }
            temp = temp.next;
        }
        //将节点接入链表，形成双向链表
        temp.next = person;
        person.pre = temp;
    }

    //添加节点，按编号顺序添加
    public void addByOrder(PersonNode person) {
        PersonNode temp = head;

        boolean flag = false; //标志添加的编号是否存在
        while (true) {
            if (temp.next == null) {
                break;
            }
            //因为是单链表，找的temp是位于添加位置的前一个节点。
            if (temp.no > person.no) {
                break;
            } else if (temp.no == person.no) {
                flag = true;
                break;
            } else {
                temp = temp.next;
            }
        }
        if (flag) {
            System.out.println("不能添加，编号已存在: " + person);
        } else {
            person.next = temp;
            person.pre = temp.pre;
            temp.pre.next = person;
            temp.pre = person;

        }
    }

    //修改链表
    public void update(PersonNode person) {
        if (head.next == null) {
            System.out.println("链表为空");
            return;
        }
        PersonNode temp = head;
        //标志有没找到对应的节点
        boolean flag = false;
        while (true) {
            if (temp == null) {
                break;
            }
            if (temp.no == person.no) {
                flag = true;
                break;
            }
            temp = temp.next;
        }
        if (flag) {
            temp.name = person.name;
            temp.nickName = person.nickName;
        } else {
            System.out.println("没有找到该编号的节点：" + person);
        }
    }

    //删除节点
    public void delete(int no) {
        if (head.next == null) {
            System.out.println("链表为空，无法删除");
            return;
        }
        PersonNode temp = head;
        boolean flag = false; //标志要删除的编号是否存在
        while (true) {
            if (temp == null) {
                break;
            }
            //因为是单链表，找的temp是位于添加位置的前一个节点。
            if (temp.no == no) {
                flag = true;
                break;
            } else if (temp.no > no) {
                break;
            } else {
                temp = temp.next;
            }
        }
        if (!flag) {
            System.out.println("不能添加，编号不存在: " + no);
        } else {
            temp.pre.next = temp.next;
            //删除最后一个节点的情况
            if(temp.next != null){
                temp.next.pre = temp.pre;
            }
        }
    }

    public static void main(String[] args) {
        PersonNode hero1 = new PersonNode(1, "宋江", "及时雨");
        PersonNode hero2 = new PersonNode(2, "卢俊义", "玉麒麟");
        PersonNode hero3 = new PersonNode(3, "吴用", "智多星");
        PersonNode hero4 = new PersonNode(4, "林冲", "豹子头");

        DoubleLinkedList doubleLinkedList = new DoubleLinkedList();
        doubleLinkedList.add(hero1);
        doubleLinkedList.add(hero2);
        doubleLinkedList.add(hero4);
        doubleLinkedList.addByOrder(hero3);
        System.out.println("----添加----");
        doubleLinkedList.list();

        hero2.name="小卢";
        doubleLinkedList.update(hero2);
        System.out.println("----更新----");
        doubleLinkedList.list();

        doubleLinkedList.delete(1);
        doubleLinkedList.delete(2);
        doubleLinkedList.delete(3);
        doubleLinkedList.delete(4);
        System.out.println("----删除----");
        doubleLinkedList.list();
    }
}


class PersonNode {
    int no;
    String name;
    String nickName;
    PersonNode next;
    PersonNode pre;

    public PersonNode(int no, String name, String nickName) {
        this.no = no;
        this.name = name;
        this.nickName = nickName;
    }

    @Override
    public String toString() {
        return "HeroNode{" +
                "no=" + no +
                ", name='" + name + '\'' +
                ", nickName='" + nickName + '\'' +
                '}';
    }
}

五、单向循环链表

5.1 约瑟夫(Josephu)问题

  1. 设编号为1,2,3...n的n个人围坐一圈，约定编号为k的人从1开始报数，数到m的那个人出列。
  2. 出列人的下一位又从1开始报数，数到m的那个人出列。
  3. 以此类推，直到所有人出列，产生一个出队编号的序列。

5.2 代码实现

package a03.linkedList;

import java.util.Arrays;

public class CircleLinkedList {
    //创建一个头结点
    private PeopleNode head;
    private int size;

    //添加节点
    public void add(PeopleNode people) {
        //如果没有节点，直接加入
        if (size == 0) {
            head = people;
            size++;
            return;
        }
        //有节点，找到最后一个加入的节点
        PeopleNode temp = head;
        while (true) {
            if (temp.next == head) {
                break;
            }
            temp = temp.next;
        }
        temp.next = people;
        people.next = head;
        size++;
    }

    //删除节点
    public void delete(int no) {
        if (size == 0) {
            System.out.println("链表为空");
            return;
        }
        //找到待删除节点的前一个节点
        PeopleNode temp = head;
        boolean flag = false;
        while (true) {
            if (temp.next.no == no) {
                flag = true;
                break;
            }
            if (temp.next == head) {
                break;
            }
            temp = temp.next;
        }
        //删除
        if (flag) {
            //如果链表只剩一个节点，头结点置为空
            if (size == 1) {
                head = null;
                size--;
                return;
            }
            //每次删除节点后，头结点改为下一个节点
            head=temp.next.next;
            temp.next = temp.next.next;
            size--;
        }
    }


    //改变指针头
    public void changeHead( int no) {
        if (head == null) {
            System.out.println("链表为空");
        }
        //找到对应节点
        PeopleNode temp = head;
        boolean flag = false;
        while (true) {
            if (temp.no == no) {
                flag = true;
                break;
            }
            if (temp.next == head) {
                break;
            }
            temp = temp.next;
        }
        //将其置为头结点
        if (flag) {
            head = temp;
        }
    }

    //显示链表
    public void show() {
        if (head == null) {
            System.out.println("链表为空");
            return;
        }
        PeopleNode temp = head;
        while (true) {
            System.out.println(temp);
            if (temp.next == head) {
                break;
            }
            temp = temp.next;
        }
    }


    //约瑟夫出圈
    public int[] outers(int k, int m) {
        int[] result = new int[size];
        int index  = 0;
        changeHead(k);
        while(true){
            if(head == null){
                break;
            }
            PeopleNode temp = head;
            for(int i=1; i <=m; i++){
                if(i == m){
                    result[index] = temp.no;
                    index++;
                    delete(temp.no);
                }
                temp = temp.next;
            }
        }
        return result;
    }

    public static void main(String[] args) {
        CircleLinkedList circleLinkedList = new CircleLinkedList();
        for (int i = 1; i <= 25; i++) {
            circleLinkedList.add(new PeopleNode(i));
        }
        System.out.println("----添加----");
        circleLinkedList.show();

//        circleLinkedList.delete(2);
//        circleLinkedList.delete(1);
//        circleLinkedList.delete(3);
//        System.out.println("----删除----");
//        circleLinkedList.show();
        System.out.println(Arrays.toString(circleLinkedList.outers(5,7)));
    }
}

class PeopleNode {
    int no;
    PeopleNode next = this;

    public PeopleNode(int no) {
        this.no = no;
    }

    @Override
    public String toString() {
        return "HeroNode{" +
                "no=" + no +
                '}';
    }
}

Q.E.D.

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