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:
Example 1:
Input: head = [4,5,1,9], node = 5 Output: [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 = 1 Output: [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:
1 The linked list will have at least two elements. 2 All of the nodes' values will be unique. 3 The given node will not be the tail and it will always be a valid node of the linked list. 4 Do not return anything from your function.
解题思路：与之前删除链表结点不同的是：这道题只给了我们要删除的那个结点，并没有给出链表的头结点，所以无法找到上一个结点，顺链删除此结点。我们可以将下一个结点的值覆盖在当前节点上，删除下一个节点即可。
C++：
1 void deleteNode(ListNode* node) { 2 ListNode* p=node->next; 3 node->val=p->val; 4 node->next=p->next; 5 delete p; 6 }
Java：
1 public void deleteNode(ListNode node) { 2 ListNode p=node.next; 3 node.val=p.val; 4 node.next=p.next; 5 }