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Formatted question description: https://leetcode.ca/all/2095.html

# 2095. Delete the Middle Node of a Linked List (Medium)

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

Similar Questions:

## Solution 1.

• /**
* 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 {
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;
}
}

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

• # 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]:
while fast and fast.next:
slow = slow.next
fast = fast.next.next
slow.next = slow.next.next
return dummy.next


• /**
* type ListNode struct {
*     Val int
*     Next *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
}

• /**
* 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 {
while (fast.next && fast.next.next) {
slow = slow.next;
fast = fast.next.next;
}
slow.next = slow.next.next;
}