2095. Delete the Middle Node of a Linked List.cpp

/*
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
*/

/**
 * 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==NULL || head->next==NULL)
            return NULL;
        
        ListNode *ptr1 = head;
        ListNode *ptr2 = head->next;
        
        while(ptr2!=NULL)
        {
            ptr2 = ptr2->next;
            
            if(ptr2!=NULL)
            {
                ptr2 = ptr2->next;
            }
            ptr1 = ptr1->next;
        }
        
        ListNode *ptr = head;
        while(ptr->next !=ptr1)
        {
            ptr = ptr->next;
        }
        ptr->next = ptr1->next;
        delete(ptr1);
        
        return head;
    }
};