Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.

Given linked list -- head = [4,5,1,9], which looks like following:

4 -> 5 -> 1 -> 9

Example 1:

Input: head = [4,5,1,9], node = 5Output: [4,1,9]Explanation: You are given the second node with value 5, the linked list             should become 4 -> 1 -> 9 after calling your function.

Example 2:

Input: head = [4,5,1,9], node = 1Output: [4,5,9]Explanation: You are given the third node with value 1, the linked list             should become 4 -> 5 -> 9 after calling your function.

Note:

• The linked list will have at least two elements.
• All of the nodes' values will be unique.
• The given node will not be the tail and it will always be a valid node of the linked list.
• Do not return anything from your function.

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对于linked list来说，当要求删除一个节点（curr）时，通常需要该节点的前一个节点（prev），使其next 指针直接指向next 节点。但这道题并没有给出前一个节点，想要删除给定的当前节点，可以通过改变当前节点的val使其变成它的下一个节点。

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         node->val = node->next->val;   //change this node's value => its next node value13         node->next = node->next->next; //skip next node14     }15 };