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. 

Have walker stop before middle node. The stop condidtion is runner != null && runner.next != null && runner.next.next != null.

Then do walker.next = walker.next.next.

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

Space: O(1).

AC Java:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode deleteMiddle(ListNode head) {
        if(head == null){
            return head;
        }
        
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode runner = dummy;
        ListNode walker = dummy;
        while(runner != null && runner.next != null && runner.next.next != null){
            runner = runner.next.next;
            walker = walker.next;
        }
        
        walker.next = walker.next.next;
        return dummy.next;
    }
}

AC C++:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* deleteMiddle(ListNode* head) {
        if(!head || !head->next){
            return NULL;
        }

        ListNode *dummy = new ListNode(-1, head);
        ListNode *walker = dummy;
        ListNode *runner = dummy;
        while(runner && runner->next && runner->next->next){
            runner = runner->next->next;
            walker = walker->next;
        }

        walker->next = walker->next->next;
        return dummy->next;
    }
};

AC Python:

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def deleteMiddle(self, head: Optional[ListNode]) -> Optional[ListNode]:
        if not head.next:
            return None
        
        dummy = ListNode(-1, head)
        walker = runner = dummy
        while(runner and runner.next and runner.next.next):
            walker = walker.next
            runner = runner.next.next

        walker.next = walker.next.next
        return dummy.next

类似Middle of the Linked List, Remove Nth Node From End of List.