[LeetCode] 2095. Delete the Middle Node of a Linked List

You are given the head of a linked list. Delete the middle node, and return the head of the modified linked list.

The middle node of a linked list of size n is the ⌊n / 2⌋th node from the start using 0-based indexing, where ⌊x⌋ denotes the largest integer less than or equal to x.

• For n = 1, 2, 3, 4, and 5, the middle nodes are 0, 1, 1, 2, and 2, respectively. 

Example 1:

Input: head = [1,3,4,7,1,2,6]
Output: [1,3,4,1,2,6]
Explanation:
The above figure represents the given linked list. The indices of the nodes are written below.
Since n = 7, node 3 with value 7 is the middle node, which is marked in red.
We return the new list after removing this node. 

Example 2:

Input: head = [1,2,3,4]
Output: [1,2,4]
Explanation:
The above figure represents the given linked list.
For n = 4, node 2 with value 3 is the middle node, which is marked in red.

Example 3:

Input: head = [2,1]
Output: [2]
Explanation:
The above figure represents the given linked list.
For n = 2, node 1 with value 1 is the middle node, which is marked in red.
Node 0 with value 2 is the only node remaining after removing node 1.

Constraints:

• The number of nodes in the list is in the range [1, 105].
• 1 <= Node.val <= 105

删除链表的中间节点。

给你一个链表的头节点 head 。删除 链表的 中间节点 ，并返回修改后的链表的头节点 head 。

长度为 n 链表的中间节点是从头数起第 ⌊n / 2⌋ 个节点（下标从 0 开始），其中 ⌊x⌋ 表示小于或等于 x 的最大整数。

对于 n = 1、2、3、4 和 5 的情况，中间节点的下标分别是 0、1、1、2 和 2 。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/delete-the-middle-node-of-a-linked-list
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这道题考察的链表的快慢指针，类似876题。唯一的不同点是876题请你返回的是链表中间的那个节点，所以走到中间那个节点之后就可以返回了。但是这道题让你返回的是去掉中间节点之后剩余的部分，返回的是头结点。虽然也是快慢指针的思路，但是这里我们还需要一个额外的 ListNode 记录 slow 指针的上一个位置。当 fast 指针跑完整个链表的时候，slow 指针指向的是中间的那个节点，但是我们需要的是 slow 节点的上一个节点，这样才能把 slow 指针指向的中间节点跳过。

时间O(n)

空间O(n)

Java实现

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode() {}
 7  *     ListNode(int val) { this.val = val; }
 8  *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 9  * }
10  */
11 class Solution {
12     public ListNode deleteMiddle(ListNode head) {
13         // corner case
14         if (head == null || head.next == null) {
15             return null;
16         }
17         
18         // normal case
19         ListNode dummy = new ListNode(0);
20         dummy.next = head;
21         ListNode fast = head;
22         ListNode slow = head;
23         ListNode prev = null;
24         while (fast != null && fast.next != null) {
25             prev = slow;
26             slow = slow.next;
27             fast = fast.next.next;
28         }
29         prev.next = prev.next.next;
30         return dummy.next;
31     }
32 }

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