C#LeetCode刷题之#237-删除链表中的节点（Delete Node in a Linked List）

问题

请编写一个函数，使其可以删除某个链表中给定的（非末尾）节点，你将只被给定要求被删除的节点。

现有一个链表 -- head = [4,5,1,9]，它可以表示为:

4 -> 5 -> 1 -> 9

输入: head = [4,5,1,9], node = 5

输出: [4,1,9]

解释: 给定你链表中值为 5 的第二个节点，那么在调用了你的函数之后，该链表应变为 4 -> 1 -> 9.

输入: head = [4,5,1,9], node = 1

输出: [4,5,9]

解释: 给定你链表中值为 1 的第三个节点，那么在调用了你的函数之后，该链表应变为 4 -> 5 -> 9.

说明:

链表至少包含两个节点。

链表中所有节点的值都是唯一的。

给定的节点为非末尾节点并且一定是链表中的一个有效节点。

不要从你的函数中返回任何结果。

Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.

Given linked list -- head = [4,5,1,9], which looks like following:

4 -> 5 -> 1 -> 9

Input: head = [4,5,1,9], node = 5

Output: [4,1,9]

Explanation: You are given the second node with value 5, the linked list should become 4 -> 1 -> 9 after calling your function.

Input: head = [4,5,1,9], node = 1

Output: [4,5,9]

Explanation: You are given the third node with value 1, the linked list should become 4 -> 5 -> 9 after calling your function.

Note:

The linked list will have at least two elements.

All of the nodes' values will be unique.

The given node will not be the tail and it will always be a valid node of the linked list.

Do not return anything from your function.

示例

public class Program {

    public static void Main(string[] args) {

        var head = new ListNode(1) {

            next = new ListNode(2) {

                next = new ListNode(3) {

                    next = new ListNode(4) {

                        next = new ListNode(5)

                    }

                }

            }

        };

        DeleteNode(head.next.next);

        ShowArray(head);

        Console.ReadKey();

    }

    private static void ShowArray(ListNode list) {

        var node = list;

        while(node != null) {

            Console.Write($"{node.val} ");

            node = node.next;

        }

        Console.WriteLine();

    }

    private static void DeleteNode(ListNode node) {

        //先复制下一个节点的值

        node.val = node.next.val;

        //再把下一节点的指针指向下下一个节点

        node.next = node.next.next;

    }

    public class ListNode {

        public int val;

        public ListNode next;

        public ListNode(int x) { val = x; }

    }

}

以上给出1种算法实现，以下是这个案例的输出结果：

1 2 4 5

分析：

显而易见，以上算法的时间复杂度为： $O(1)$ 。

巴特西

C#LeetCode刷题之#237-删除链表中的节点（Delete Node in a Linked List）

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