The sword refers to Offer-question 18 (Java Edition): delete linked list nodes in O(1) time

Reference from: "Sword Pointing Offer - Famous Enterprise Interviewer Talking About Typical Programming Questions"

Title : Deleting a Linked List Node in O(1) Time
Given the head pointer of a singly linked list and a node pointer, define a function to delete the node in O(1) time.

Main idea :
There are three cases:
1. The node to be deleted is not the tail node: replace the node to be deleted with its next node.
2. The linked list has only one node, that is, the head node is the node to be deleted: directly set the head node to null.
3. The node to be deleted is the tail node: find the previous node of the node to be deleted, and set the next node of the found node to null.

Key points : node replacement, special nodes (head node and tail node)

Time Complexity : O(1) on average

public class DeleteNodeInList
{
    public static void main(String[] args)
    {
        int[] data = {3, 4, 5, 7};
        ListNode pHead = LinkedListData.GenerateDataByArray(data);
        ListNode deletedNode5 = pHead.next.next; //5
        ListNode deletedNode7 = pHead.next.next.next; //7
        ListNode result = deleteNode(pHead, pHead);
        while (result != null)
        {
            System.out.println(result.val);
            result = result.next;
        }
    }

    private static ListNode deleteNode(ListNode head, ListNode deletedNode)
    {
        if (head == null || deletedNode == null) return null;
        //删除的结点不是尾结点
        if (deletedNode.next != null)
        {
            //将要删除的结点替换成它的下一个结点
            ListNode nextNode = deletedNode.next;
            deletedNode.val = nextNode.val;
            deletedNode.next = nextNode.next;
        }
        //链表只有一个结点，即头结点就是要删除的结点
        else if (head.val == deletedNode.val)
        {
            head = null;
        }
        //删除的结点是尾结点
        else
        {
            ListNode pNode = head;
            //找到要删除的结点的前一个结点
            while (pNode.next.val != deletedNode.val)
            {
                pNode = pNode.next;
            }
            pNode.next = null;
        }
        return head;
    }
}

class ListNode
{
    int val;
    ListNode next = null;

    ListNode(int val)
    {
        this.val = val;
    }
}

class LinkedListData
{
     /**
     * 根据数组生成链表
     * @param data
     * @return
     */
public static ListNode GenerateDataByArray(int[] data)
    {
        ListNode head = new ListNode(data[0]);
        ListNode currentNode = head;
        for (int i = 1; i < data.length; i++)
        {
            ListNode next = new ListNode(data[i]);
            currentNode.next = next;
            currentNode = next;
        }
        return head;
    }
}