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

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)

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