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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.

// OJ: https://leetcode.com/problems/delete-the-middle-node-of-a-linked-list/
// Time: O(N)
// Space: O(1)
class Solution {
int getLength(ListNode *head) {
int ans = 0;
for (; head; ++ans, head = head->next);
return ans;
}
public:
ListNode* deleteMiddle(ListNode* head) {
ListNode dummy, *p = &dummy;
dummy.next = head;
int len = getLength(head) / 2;
for (int i = 0; i < len; ++i, p = p->next);
p->next = p->next->next;
return dummy.next;
}
};


## Solution 2. Fast-Slow Pointers

// OJ: https://leetcode.com/problems/delete-the-middle-node-of-a-linked-list/
// Time: O(N)
// Space: O(1)
class Solution {
public:
ListNode* deleteMiddle(ListNode* head) {
ListNode dummy, *fast = &dummy, *slow = &dummy;
dummy.next = head;
while (fast->next && fast->next->next) {
slow = slow->next;
fast = fast->next->next;
}
slow->next = slow->next->next;
return dummy.next;
}
};