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.
题目不难，删除节点，但是方法很巧，将后一个节点直接覆盖当前节点，不需要头指针也能删除节点，唯一不足是不能删除最后一个节点
代码如下：
1 /**
2 * Definition for singly-linked list.
3 * struct ListNode {
4 * int val;
5 * ListNode *next;
6 * ListNode(int x) : val(x), next(NULL) {}
7 * };
8 */
9 class Solution {
10 public:
11 void deleteNode(ListNode* node) {
12 ListNode *tmp = node->next;
13 node->val = tmp->val;
14 node->next = tmp->next;
15 delete tmp;
16 }
17 };
时间复杂度：O(1)
空间复杂度：O(1)
