LeetCode 237. Delete Node in a Linked List

最新推荐文章于 2025-04-19 00:22:55 发布

原创最新推荐文章于 2025-04-19 00:22:55 发布 · 263 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode

算法专栏收录该内容

30 篇文章

订阅专栏

本文介绍了一个C++函数，用于删除给定链表中除尾节点外的任意节点。该方法通过复制后继节点的值并调整指针来实现，最终使待删除节点从链表中移除。

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.

这题自己A了好久没过

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    void deleteNode(ListNode* node) {
        ListNode* cur = node;
        ListNode* post = cur->next; // except the tail, thus has access to the next
        while(true)
        {
            cur->val = post->val;
            if(post->next == NULL)
            {
                cur->next = NULL;
                return;
            }
            else
            {
                cur = post;
                post = post->next;
            }
        }
    }
};