2095. Delete the Middle Node of a Linked List|Leetcode solution

 2095. Delete the Middle Node of a Linked List

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

solution:

/**

 * 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->next == NULL )

            return NULL;

        if(head ->next->next == NULL)

        {

            head->next = NULL;

            return head ;

        }

        

        ListNode* slow = head;

        ListNode* fast = head;

        while(fast->next->next != NULL and fast->next->next->next != NULL)

        {

            slow = slow->next;

            fast = fast->next->next;

        }

        slow ->next = slow->next->next;

        return head;

        

    }

};