LintCode: Delete Node in the Middle of Singly Linked List

开始没看懂题目的意思，以为是输入一个单链表，删掉链表中间的那个节点。

实际的意思是，传入的参数就是待删节点，所以只要把当前节点指向下一个节点就可以了。

C++

 /**
  * Definition of ListNode
  * class ListNode {
  * public:
  *     int val;
  *     ListNode *next;
  *     ListNode(int val) {
  *         this->val = val;
  *         this->next = NULL;
  *     }
  * }
  */
 class Solution {
 public:
     /**
      * @param node: a node in the list should be deleted
      * @return: nothing
      */
     void deleteNode(ListNode *node) {
         // write your code here
         node->val = node->next->val;
         node->next = node->next->next;
     }
 };