LeetCode 237 Delete Node in a Linked List（在链表中删除节点）

翻译

给定一个访问节点的路径，写一个函数去删除在一个单向链表中除尾部以外的节点。

假设这个链表是1 -> 2 -> 3 -> 4，并且你被给予了第3个值为3的节点，那么在调用你的函数之后这个链表应该变为1 -> 2 -> 4。

原文

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.

代码

/**
* 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) {
        if (node == NULL) {}
        else {
            ListNode* tmp = node->next;
            node->val = tmp->val;
            node->next = tmp->next;
            delete tmp;
        }
    }
};