Given a singly linked list, delete middle of the linked list. For example, if given linked list is 1->2->3->4->5 then linked list should be modified to 1->2->4->5.

If there are even nodes, then there would be two middle nodes, we need to delete the second middle element. For example, if given linked list is 1->2->3->4->5->6 then it should be modified to 1->2->3->5->6.

If the input linked list is NULL or has 1 node, then it should return NULL

Example 1:

Input:

Output: 1 2 4 5

Example 2:

Input:

Output: 2 4 6 5 1

create two pointers: fast and slow.

point both of them to head at first.

Now, slow moves by one position and fast moves by two position.

when fast points to the end of the circular linked list slow will point the middle element of the list

now delete this slow element

c++ implementation:

`Node* deleteMid(Node* head){    if(head == NULL)    {      return head;    }    Node *slow = head, *fast = head, *prev = head;    while(fast != NULL && fast->next != NULL)    {        prev = slow;        slow = slow->next;        fast = fast->next->next;    }    prev->next = slow->next;    return head;}`
Time Complexity: O(n)

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