LeetCode 2095. Delete the Middle Node of a Linked List

原题链接在这里：https://leetcode.com/problems/delete-the-middle-node-of-a-linked-list/

题目：

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

题解：

Find the pre node of middle node. 

Then do pre.next = pre.next.next.

Time Complexity: O(n). n is the length of list.

Space: O(1).

AC Java:

 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         if(head == null || head.next == null){
14             return null;
15         }
16         
17         ListNode walker = head;
18         ListNode runner = head;
19         ListNode pre = null;
20         while(runner != null && runner.next != null){
21             pre = walker;
22             walker = walker.next;
23             runner = runner.next.next;
24         }
25         
26         pre.next = pre.next.next;
27         return head;
28     }
29 }

类似.