LeetCode OJ：Delete Node in a Linked List（链表节点删除）-优快云博客

本文提供了一种方法，在不修改原始链表结构的情况下，通过替换当前节点值来删除单链表中的指定节点。示例代码演示了如何在Java环境下实现这一功能。

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

Supposed the linked list is 1 -> 2 -> 3 -> 4 and you are given the third node with value 3, the linked list should become 1 -> 2 -> 4 after calling your function.

更改当前节点的值来实现

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 
10 class Solution {
11 public:
12     void deleteNode(ListNode* node) {
13         if(node == NULL) return;
14         node->val = node->next->val;
15         node->next = node->next->next;
16     }
17 };

没什么好说的，下面是java版本的：

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) { val = x; }
 7  * }
 8  */
 9 public class Solution {
10     public void deleteNode(ListNode node) {
11         node.val = node.next.val;
12         node.next = node.next.next;
13     }
14 }