June Leetcoding Challenge(6.2)

Delete Node in a Linked List

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：

由于题中给出的方法的参数中只提供了要删除节点，因此，只要将当前节点的value值改为下一个节点的value，将当前节点的next指向下一个节点的next。即例子中删除node=5，将5节点的value值改为1，节点next值指向节点9。

代码：

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