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.

问题:只给定单向列表中的一个节点,从列表中删除该节点。

删除列表的某个元素,只知道根据头节点,和待删节点值,遍历搜索到相同值,将其删除。对于给定节点从列表中删除,没有思路。在网上看了讲解,原来这个是列表的基本操作,O(1) 时间即可完成。看了我对列表还不够熟悉。

     void deleteNode(ListNode* node) {

         node->val = node->next->val;
node->next = node->next->next; }

题目提到给定的节点,不会是最末尾节点。这样看了这个算法是有缺陷的,因为不能删除最后一个元素。其实不是,只需要在原本的列表最后,插入一个符号代表 NULL ,那么这个算法就对整个列表的完整操作。

参考资料:

[LeetCode]Delete Node in a Linked List, 书影

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