237. Delete Node in a Linked List

最新推荐文章于 2025-09-07 20:32:33 发布

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原文链接：http://www.cnblogs.com/chucklu/p/10509969.html

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#数据结构与算法

本文介绍了一种在单链表中删除指定节点（除尾节点）的方法，无需访问头节点。通过将目标节点的值替换为其后继节点的值，然后删除最后一个节点，实现了在仅给定待删除节点的情况下进行链表节点的删除操作。

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:

Example 1:

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.

Example 2:

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.

刚开始题目都没看懂，后来看了discuss才知道，题目想要做什么。

给定链表中的一个结点，把这个结点删除。因为只是给定了结点，并没有链表的head。

所以，处理方式是，将结点的值，往前移动，然后删除最后一个结点。

public void DeleteNode(ListNode node)
        {
            ListNode nodePrev = node;
            while (node.next != null)
            {
                node.val = node.next.val;
                nodePrev = node;
                node = node.next;
            }

            nodePrev.next = null;
        }