Data Structure: What is the efficient way to delete middle node of a singly linked list ?

Question

How to delete the node that given as argument in single linked list?
Is there any system call to get prev pointer ?

Note: Here the goal is to delete the address not the node's value

Answer 1

This jobs can be done once you are able to find out the middle node of the Link List. Assuming the link list is single link list we can follow the following algo -
1. Take two pointers slow_ptr and fast_ptr.
2. make fast_ptr move two nodes in slow_ptr moves one node in each iteration more than slow *ptr.
3. when fast_ptr reaches to null then slow_ptr will point to middle node and you can delete it.

Void deleteMidddle(Node* head)
{
     Node* slow_ptr = head;
     Node* fast_ptr = head;
     Node* tmp = head;
     while(slow_ptr->next != NULL && fast_ptr->next != NULL)
     {
        tmp = slow_ptr;
        slow_ptr = slow_ptr->next;
        fast_ptr = fast_ptr->next->next;
     }
     tmp->next = slow_ptr->next;
     free(slow_ptr);
}

Answer 2

1.assign a fast *ptr and slow *ptr.
2.make fast *ptr to move 2nodes more than slow *ptr..
3.when fast *ptr ends to last node,then we see slow *ptr wiil be at middle..

Answer 3

Void deleteMidddleInLL(Node* start)
{
     Node* slow_ptr = start;
     Node* fast_ptr = start;
     Node* temp = start;
     while(slow_ptr->next != NULL && fast_ptr->next != NULL)
     {
        temp = slow_ptr;
        slow_ptr = slow_ptr->next;
        fast_ptr = fast_ptr->next->next;
     }
     temp->next = slow_ptr->next;
     free(slow_ptr);
}