【算法与数据结构】单链表的增删改查、逆序打印与输出、合并有序链表

最近博主在B站学习算法与数据结构，视频链接：

https://www.bilibili.com/video/BV1E4411H73v?p=23

这是一道课后练习，题目是：合并两个有序的单链表，使合并后的链表依然有序。

代码如下，合并部分的代码是mergeTwoSingleLinkedList：

import java.util.Stack;

public class SingleLinkedListDemo{

    public static void main(String[] args){
        boolean flag = false;  //　true=直接添加
        HeroNode player1 = new HeroNode(23,"詹姆斯","King");
        HeroNode player2 = new HeroNode(24,"科比","黑曼巴");
        HeroNode player3 = new HeroNode(2,"韦德","闪电侠");
        HeroNode player4 = new HeroNode(3,"戴维斯","浓眉");

        HeroNode newPlayer1 = new HeroNode(14,"丹尼格林","皇阿玛");
        HeroNode newPlayer2 = new HeroNode(32,"麦基","囧哥");
        HeroNode newPlayer3 = new HeroNode(34,"阿德托昆博","字母哥");
        HeroNode newPlayer4 = new HeroNode(0,"维斯布鲁克","神龟");
        SingleLinkedList singleLinkedList = new SingleLinkedList();
        SingleLinkedList newsingleLinkedList = new SingleLinkedList();
        if (flag) {
           singleLinkedList.add(player1);
           singleLinkedList.add(player2);
           singleLinkedList.add(player3);
           singleLinkedList.add(player4);
           System.out.println("直接添加后的链表情况:");
           singleLinkedList.list();
           System.out.println("
");
        }else {
           singleLinkedList.addByOrder(player1);
           singleLinkedList.addByOrder(player2);
           singleLinkedList.addByOrder(player3);
           singleLinkedList.addByOrder(player4);
           System.out.println("链表1排序后的链表情况:");
           singleLinkedList.list();
           newsingleLinkedList.addByOrder(newPlayer1);
           newsingleLinkedList.addByOrder(newPlayer2);
           newsingleLinkedList.addByOrder(newPlayer3);
           newsingleLinkedList.addByOrder(newPlayer4);
           System.out.println("链表2排序后的链表情况：");
           newsingleLinkedList.list();
        }
        //merge two singleLinkedList
        System.out.println("合并两个有序链表并返回一个有序链接：");
        mergeTwoSingleLinkedList(singleLinkedList.getHead(),newsingleLinkedList.getHead());
        singleLinkedList.list();
        System.out.println("***************----------------***************");
        //update
        HeroNode newPlayer = new HeroNode(2,"欧文","小王爷");
        singleLinkedList.update(newPlayer);
        System.out.println("修改后的链表情况:");
        singleLinkedList.list();
        //delete
        singleLinkedList.del(24);
        System.out.println("删除后的链表情况:");
        singleLinkedList.list();
        //查找倒数第k个节点
        HeroNode res = findLastIndexNode(singleLinkedList.getHead(),2);
        System.out.println("倒数第k个节点res="+ res);
        //求单链表有效节点的个数
        System.out.println("有效的节点个数="+getLength(singleLinkedList.getHead()));
        //逆序打印单链表，不改变链表结构
        System.out.println("逆序打印单链表，不改变单链表的结构");
        reversePrint(singleLinkedList.getHead());
        System.out.println("原链表如下：");
        singleLinkedList.list();
        //反转单链表
        System.out.println("反转单链表");
        reverseList(singleLinkedList.getHead());
        singleLinkedList.list();
    }

    public static void mergeTwoSingleLinkedList(HeroNode head1, HeroNode head2){
        if (head1.next == null || head2.next == null){
            return;//若其中1个链表为空，不进行合并
        }
        HeroNode temp = head1;  // 指向head1链表的头节点
        HeroNode cur2 = head2.next;  // 指向head2链表头节点的下一个节点
        HeroNode next1 = null;  //保存temp的下一个节点
        HeroNode next2 = null;  //保存cu2的下一个节点
        while(cur2 != null){
            next2 = cur2.next;
            while (temp.next != null){
                next1 = temp.next;
                if (cur2.no <= temp.next.no){  //当cur2的no小于等于temp下一个节点的no时
                    cur2.next = temp.next; // 将cur2插入到head1中
                    temp.next = cur2;
                    break;
                }else {
                    temp = next1;  //将链表head1向后移，遍历head1
                    if(temp.next == null){// head1到最后时，将cur2添加到head1的末尾
                        temp.next = cur2;
                        cur2.next = null;
                        break;
                    }
                }
            }
            cur2 = next2;  //将head2向后移，重新遍历head1的整个链表
        }
    }
    public static void reversePrint(HeroNode head){
        if (head.next == null){
            return;
        }
        Stack<HeroNode> stack = new Stack<HeroNode>();
        HeroNode cur = head.next;
        while (cur != null){
            stack.push(cur);
            cur = cur.next;
        }
        while (stack.size()>0){
            System.out.println(stack.pop());
        }
    }
    public static void reverseList(HeroNode head){
        if (head.next == null || head.next.next == null){
            return;
        }
        HeroNode cur = head.next;
        HeroNode next = null;
        HeroNode reverseHead = new HeroNode(0,"","");
        while (cur != null){
            next = cur.next;
            cur.next = reverseHead.next;
            reverseHead.next = cur;
            cur = next;
        }
        head.next = reverseHead.next;
    }

    public static HeroNode findLastIndexNode(HeroNode head, int index){
        if(head.next == null){
            return null;
        }
        int size = getLength(head);
        if(index <=0 || index > size){
            return null;
        }
        HeroNode cur = head.next;
        for (int i=0;i<size-index;i++){
            cur = cur.next;
        }
        return cur;

    }
    public static int getLength(HeroNode head){
        if(head.next == null){
            return 0;
        }
        int length = 0;
        HeroNode cur = head.next;
        while (cur != null){
            length++;
            cur = cur.next;
        }
        return length;
    }
}

class SingleLinkedList{
    private HeroNode head = new HeroNode(0, "", "");
    public HeroNode getHead(){return head;}

    public void add(HeroNode heroNode){
        HeroNode temp = head;
        while(true){
            if(temp.next == null){
                break;
            }
            temp = temp.next;
        }
        temp.next = heroNode;
    }

    public void addByOrder(HeroNode heroNode){
        HeroNode temp = head;
        boolean flag = false;
        while(true){
            if(temp.next == null){
                break;
            }
            if(temp.next.no > heroNode.no){
                break;
            }else if (temp.next.no == heroNode.no){
                flag = true;
                break;
            }
            temp = temp.next;
        }
        if(flag){
            System.out.printf("准备插入的英雄编号%d已存在，不能加入",heroNode.no);
        } else {
            heroNode.next = temp.next;
            temp.next = heroNode;
        }
    }

    public void update(HeroNode newHeroNode){
        if(head.next == null){
            System.out.println("链表为空");
            return;
        }
        HeroNode temp = head.next;
        boolean flag = false;
        while (true){
            if (temp == null){
                break;
            }
            if(temp.no == newHeroNode.no){
                flag = true;
                break;
            }
            temp = temp.next;
        }
        if (flag){
            temp.name = newHeroNode.name;
            temp.nickname = newHeroNode.nickname;
        }else {
            System.out.printf("没有找到编号%d的节点，不能修改
", newHeroNode.no);
        }
    }

    public void del(int no){
        HeroNode temp = head;
        boolean flag = false;
        while (true){
            if (temp.next == null){
                System.out.println("已遍历完整个链表
");
                break;
            }
            if (no == temp.next.no){
                flag = true;
                break;
            }
            temp = temp.next;
        }
        if(flag){
            temp.next = temp.next.next;
        }else {
            System.out.printf("要删除的节点%d不存在
", no);
        }
    }

    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;
            }
            System.out.println(temp);
            temp = temp.next;
        }
    }
}

class HeroNode{
    public int no;
    public String name;
    public String nickname;
    public 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 + "]";
    }
}

打印结果如下：

链表1排序后的链表情况:
HeroNode [no2, name=韦德, nickname=闪电侠]
HeroNode [no3, name=戴维斯, nickname=浓眉]
HeroNode [no23, name=詹姆斯, nickname=King]
HeroNode [no24, name=科比, nickname=黑曼巴]
链表2排序后的链表情况：
HeroNode [no0, name=维斯布鲁克, nickname=神龟]
HeroNode [no14, name=丹尼格林, nickname=皇阿玛]
HeroNode [no32, name=麦基, nickname=囧哥]
HeroNode [no34, name=阿德托昆博, nickname=字母哥]
合并两个有序链表并返回一个有序链接：
HeroNode [no0, name=维斯布鲁克, nickname=神龟]
HeroNode [no2, name=韦德, nickname=闪电侠]
HeroNode [no3, name=戴维斯, nickname=浓眉]
HeroNode [no14, name=丹尼格林, nickname=皇阿玛]
HeroNode [no23, name=詹姆斯, nickname=King]
HeroNode [no24, name=科比, nickname=黑曼巴]
HeroNode [no32, name=麦基, nickname=囧哥]
HeroNode [no34, name=阿德托昆博, nickname=字母哥]
***************----------------***************
修改后的链表情况:
HeroNode [no0, name=维斯布鲁克, nickname=神龟]
HeroNode [no2, name=欧文, nickname=小王爷]
HeroNode [no3, name=戴维斯, nickname=浓眉]
HeroNode [no14, name=丹尼格林, nickname=皇阿玛]
HeroNode [no23, name=詹姆斯, nickname=King]
HeroNode [no24, name=科比, nickname=黑曼巴]
HeroNode [no32, name=麦基, nickname=囧哥]
HeroNode [no34, name=阿德托昆博, nickname=字母哥]
删除后的链表情况:
HeroNode [no0, name=维斯布鲁克, nickname=神龟]
HeroNode [no2, name=欧文, nickname=小王爷]
HeroNode [no3, name=戴维斯, nickname=浓眉]
HeroNode [no14, name=丹尼格林, nickname=皇阿玛]
HeroNode [no23, name=詹姆斯, nickname=King]
HeroNode [no32, name=麦基, nickname=囧哥]
HeroNode [no34, name=阿德托昆博, nickname=字母哥]
倒数第k个节点res=HeroNode [no32, name=麦基, nickname=囧哥]
有效的节点个数=7
逆序打印单链表，不改变单链表的结构
HeroNode [no34, name=阿德托昆博, nickname=字母哥]
HeroNode [no32, name=麦基, nickname=囧哥]
HeroNode [no23, name=詹姆斯, nickname=King]
HeroNode [no14, name=丹尼格林, nickname=皇阿玛]
HeroNode [no3, name=戴维斯, nickname=浓眉]
HeroNode [no2, name=欧文, nickname=小王爷]
HeroNode [no0, name=维斯布鲁克, nickname=神龟]
原链表如下：
HeroNode [no0, name=维斯布鲁克, nickname=神龟]
HeroNode [no2, name=欧文, nickname=小王爷]
HeroNode [no3, name=戴维斯, nickname=浓眉]
HeroNode [no14, name=丹尼格林, nickname=皇阿玛]
HeroNode [no23, name=詹姆斯, nickname=King]
HeroNode [no32, name=麦基, nickname=囧哥]
HeroNode [no34, name=阿德托昆博, nickname=字母哥]
反转单链表
HeroNode [no34, name=阿德托昆博, nickname=字母哥]
HeroNode [no32, name=麦基, nickname=囧哥]
HeroNode [no23, name=詹姆斯, nickname=King]
HeroNode [no14, name=丹尼格林, nickname=皇阿玛]
HeroNode [no3, name=戴维斯, nickname=浓眉]
HeroNode [no2, name=欧文, nickname=小王爷]
HeroNode [no0, name=维斯布鲁克, nickname=神龟]