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.

题意很简单很简单，删除单链表中指定的结点，仔细一看，不对啊。。无法得到要删除结点的前驱结点，也就是说不能用传统的方法。如果单链表1->2->3->4->5,我们要删除结点3，可以让3赋值为4，3指向5。那么就变成了1->2->4->5，^_^结果就得到了

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     struct ListNode *next; * }; */void deleteNode(struct ListNode* node) {    if(node==NULL)    {        return ;    }    else    {        node->val=node->next->val;        node->next=node->next->next;    }}