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.

 

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一上来就感觉这个题没法做啊，没有提供头结点，只是被删除的节点，怎么找到被删除节点的上一个节点啊。

看了下其他人的答案，才发现自己已经形成了思维定势。

也可以采用节点内赋值的方法

void deleteNode(ListNode* node) {    if (node == nullptr)        return;    node->val = node->next->val;    node->next = node->next->next;}