LeetCode 237. Delete Node in a Linked List 删除链表结点（只给定要删除的结点） C++/Java

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     }