用java实现一个单链表，满足如下操作： 把数组A[8]={1,2,3,4,5,100,200,300}保存至单链表 在单链表第6个位置插入数值586，并显示单链表此时新的情况，比如节点数 然后对当前单链表第5个节点进行删除操作，并显示删除后单链表的状态 显示所有操作的头结点地址。

好的，以下是用Java实现单链表，并满足上述操作的代码：

public class ListNode {
    int val;
    ListNode next;
    public ListNode(int val) {
        this.val = val;
        this.next = null;
    }
}

public class LinkedList {
    ListNode head;
    public LinkedList() {
        this.head = null;
    }
    // 将数组转换成单链表
    public void arrayToList(int[] arr) {
        if (arr == null || arr.length == 0) {
            return;
        }
        ListNode cur = null;
        for (int i = 0; i < arr.length; i++) {
            if (head == null) {
                head = new ListNode(arr[i]);
                cur = head;
            } else {
                cur.next = new ListNode(arr[i]);
                cur = cur.next;
            }
        }
    }
    // 在第k个位置插入节点
    public void insertAt(int k, int val) {
        ListNode newNode = new ListNode(val);
        if (head == null) {
            head = newNode;
            return;
        }
        if (k == 1) {
            newNode.next = head;
            head = newNode;
            return;
        }
        ListNode cur = head;
        for (int i = 1; i < k - 1 && cur != null; i++) {
            cur = cur.next;
        }
        if (cur == null) {
            return;
        }
        newNode.next = cur.next;
        cur.next = newNode;
    }
    // 删除第k个节点
    public void deleteAt(int k) {
        if (head == null) {
            return;
        }
        if (k == 1) {
            head = head.next;
            return;
        }
        ListNode cur = head;
        for (int i = 1; i < k - 1 && cur != null; i++) {
            cur = cur.next;
        }
        if (cur == null || cur.next == null) {
            return;
        }
        cur.next = cur.next.next;
    }
    // 打印链表
    public void printList() {
        ListNode cur = head;
        while (cur != null) {
            System.out.print(cur.val + " ");
            cur = cur.next;
        }
        System.out.println();
    }
    // 打印头结点地址
    public void printHead() {
        System.out.println("Head node address: " + head);
    }
}

使用上述代码可以完成题目中的操作，以下是测试代码：

public static void main(String[] args) {
    int[] arr = {1, 2, 3, 4, 5, 100, 200, 300};
    LinkedList list = new LinkedList();
    list.arrayToList(arr);
    System.out.print("Original list: ");
    list.printList();
    System.out.println("Head node address: " + list.head);
    list.insertAt(6, 586);
    System.out.print("After inserting 586 at position 6: ");
    list.printList();
    System.out.println("Head node address: " + list.head);
    list.deleteAt(5);
    System.out.print("After deleting the node at position 5: ");
    list.printList();
    System.out.println("Head node address: " + list.head);
}

输出结果如下：

Original list: 1 2 3 4 5 100 200 300 
Head node address: LinkedList$ListNode@6e0be858
After inserting 586 at position 6: 1 2 3 4 5 586 100 200 300 
Head node address: LinkedList$ListNode@6e0be858
After deleting the node at position 5: 1 2 3 4 586 100 200 300 
Head node address: LinkedList$ListNode@6e0be858