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237. Delete Node in a Linked List
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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：
删除链表指定节点。前提是只传删除节点给函数。但不包括删除尾节点。
思路：
由于没有链表前节点的存在，所以删除链表时无法改变前节点的指向。但是链表值是int型的，所有可以把当前节点的val和下个节点的val交换。然后删除下个节点即可。
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
void deleteNode(struct ListNode* node)
{
if ( node->next == NULL )
{
return;
}
int tmp = 0;
tmp = node->val;
node->val = node->next->val;
node->next->val = tmp;
struct ListNode *list = node->next;
node->next = node->next->next;
free(list);
}