删除链表中的节点 Delete Node in a Linked List

题目

请编写一个函数，使其可以删除某个链表中给定的（非末尾）节点，你将只提供要求被删除的节点

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.

思路&题解

链表的基本操作，替换并指向下一个节点即可。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public void deleteNode(ListNode node) {
    ListNode next = node.next;
    node.val = next.val;
    node.next = next.next;
}