【数据结构与算法(java)】单链表 +倒数第N个节点+反转链表+逆序打印

链表

• 分为带头节点的链表和没有头结点的链表

单链表

//(带头结点)
public class SingleLinkedListDemo {

    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");

        SingleLinkedList singleLinkedList = new SingleLinkedList();
        // singleLinkedList.add(hero1);
        // singleLinkedList.add(hero2);
        // singleLinkedList.add(hero3);
        // singleLinkedList.add(hero4);

        singleLinkedList.addByOrder(hero4);
        singleLinkedList.addByOrder(hero1);
        singleLinkedList.addByOrder(hero2);
        singleLinkedList.addByOrder(hero3);

        HeroNode hero5 = new HeroNode(4, "Lin", "豹子头");

        singleLinkedList.update(hero5);
        singleLinkedList.del(1);
        singleLinkedList.list();
    }
}

class SingleLinkedList {

    private HeroNode head = new HeroNode(0, "", "");

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

    public void list() {
        if (head.next == null) {
            return;
        }
        HeroNode temp = head.next;
        while (temp != null) {
            System.out.println(temp);
            temp = temp.next;
        }
    }

    public void addByOrder(HeroNode heroNode) {
        HeroNode temp = head;
        while (temp.next != null) {
            if (temp.next.no > heroNode.no) {
                break; 
            } else if (temp.next.no == heroNode.no) {
                throw new RuntimeException("编号已存在");
            }
            temp = temp.next;
        }
        heroNode.next = temp.next;
        temp.next = heroNode;
    }

    public void update(HeroNode newHeroNode) {
        if (head.next == null) {
            return;
        }
        HeroNode temp = head.next;
        boolean flag = false;
        while (temp != null) {
            if (temp.no == newHeroNode.no) {
                temp.name = newHeroNode.name;
                temp.nickname = newHeroNode.nickname;
                flag = true;
                break;
            }
            temp = temp.next;
        }
        if (!flag) {
            throw new RuntimeException("找不到该编号的节点");
        }
    }

    public void del(int no) {
        HeroNode temp = head;
        boolean flag =false;
        while (temp.next!=null) {
            if (temp.next.no==no) {
                flag=true;
                temp.next=temp.next.next;
            }
            temp=temp.next;
        }
        if (!flag) {
            System.out.println("不存在该节点");
        }
    }
}

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 + "]";
    }
}

• 倒数第N个节点

      public static HeroNode findLastIndexNode(HeroNode head, int index) {
          if (head.next == null) {
              return null;
          }
          int length = getLength(head);
          if (index <= 0 || index > length) {
              return null;
          }
          HeroNode temp = head.next;
          int currentIndex = 0;
          while (temp != null) {
              currentIndex++;
              if (currentIndex == length + 1 - index) {
                  return temp;
              }
              temp = temp.next;
          }
          return null;
      }
  
  

• 反转链表

  思路:

  1. 先定义一个节点 reverseHead = new HeroNode();
  2. 从头到尾遍历原来的链表，每遍历一个节点，就将其取出，并放在新的链表 reverseHead 的最前端.
  3. 原来的链表的 head.next = reverseHead.next

    public static void reverseList(HeroNode head) {
        if (head.next == null || head.next.next == null) {
            return;
        }
        HeroNode newHead = new HeroNode(0, "", "");
        HeroNode cur = head.next;
        HeroNode next;
        while (cur != null) {
            next = cur.next;
            cur.next = newHead.next;
            newHead.next = cur;
            cur = next;
        }
        head.next = newHead.next;
    }

• 逆序打印链表

  思路：

  1. 法一：先反转，再遍历（但是会破坏原有结构）

  2. 法二：可以利用栈（建议）

         public static void reversePrint(HeroNode head) {
             if (head.next==null) {
                 return;
             }
             Stack<HeroNode> stack = new Stack<>();
             HeroNode temp = head.next;
             while (temp!=null) {
                 stack.push(temp);
                 temp=temp.next;
             }
             while(!stack.empty()) {
                 System.out.println(stack.pop());
             }
         }