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 value3, the linked list should become 1 -> 2 -> 4 after calling your function.
从链表中，删除指定的节点，但又没有给出前指针，，所以先交换当前节点和下一节点的值，然后删除下一节点即可。代码如下：
void deleteNode(ListNode* node) {if (!node || !node->next)return;node->val = node->next->val;node->next = node->next->next;}
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让我们将事前的忧虑，换为事前的思考和计划吧!
