README_EN.md

2095. Delete the Middle Node of a Linked List

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Description

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

Solutions

Python3

# 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]:
        dummy = ListNode(next=head)
        slow, fast = dummy, head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
        slow.next = slow.next.next
        return dummy.next

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) {
        ListNode dummy = new ListNode(0, head);
        ListNode slow = dummy, fast = head;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        slow.next = slow.next.next;
        return dummy.next;
    }
}

TypeScript

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function deleteMiddle(head: ListNode | null): ListNode | null {
    if (!head || !head.next) return null;
    let fast = head.next, slow = head;
    while (fast.next && fast.next.next) {
        slow = slow.next;
        fast = fast.next.next;
    }
    slow.next = slow.next.next;
    return head;
};

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) {
        ListNode* dummy = new ListNode(0, head);
        ListNode* slow = dummy;
        ListNode* fast = head;
        while (fast && fast->next)
        {
            slow = slow->next;
            fast = fast->next->next;
        }
        slow->next = slow->next->next;
        return dummy->next;
    }
};

Go

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func deleteMiddle(head *ListNode) *ListNode {
	dummy := &ListNode{Val: 0, Next: head}
	slow, fast := dummy, dummy.Next
	for fast != nil && fast.Next != nil {
		slow, fast = slow.Next, fast.Next.Next
	}
	slow.Next = slow.Next.Next
	return dummy.Next
}

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