[Java]LeetCode237. 删除链表中的节点 | 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.

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:

• The linked list will have at least two elements.
• All of the nodes' values will be unique.
• The given node will not be the tail and it will always be a valid node of the linked list.
• Do not return anything from your function.

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请编写一个函数，使其能够删除某个链表中给定的（非末尾）节点，你将只被给定要求被删除的节点。

现有一个链表 -- head = [4,5,1,9]，它能够表示为:

    4 -> 5 -> 1 -> 9

示例 1:

输入: head = [4,5,1,9], node = 5
输出: [4,1,9]
解释: 给定你链表中值为 5 的第二个节点，那么在调用了你的函数以后，该链表应变为 4 -> 1 -> 9.

示例 2:

输入: head = [4,5,1,9], node = 1
输出: [4,5,9]
解释: 给定你链表中值为 1 的第三个节点，那么在调用了你的函数以后，该链表应变为 4 -> 5 -> 9.

说明:

• 链表至少包含两个节点。
• 链表中全部节点的值都是惟一的。
• 给定的节点为非末尾节点而且必定是链表中的一个有效节点。
• 不要从你的函数中返回任何结果。

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0ms

1 class Solution {
2     public void deleteNode(ListNode node) {
3         node.val=node.next.val;
4         node.next=node.next.next;
5     }
6 }

---

34180 kb

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) { val = x; }
 7  * }
 8  */
 9 class Solution {
10     public void deleteNode(ListNode node) {
11         
12         int temp;
13         temp = node.val;
14         node.val = node.next.val;
15         node.next.val =  temp;
16         
17         node.next = node.next.next;        
18         
19     }
20 }