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;
    }
};