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have two pointers, each time go one step forward on the first pointer and two on the second pointer... when the second pointer reaches end of the list, first one is at the middle of the list

have two pointers, each time go one step forward on the first pointer and two on the second pointer... when the second pointer reaches end of the list, first one is at the middle of the list

ListCount is the number of elements in the list. Divide ListCount by 2. Keep start pointer in head. set another startCount variable to 0. Increment startCount by one and reach the next node till start count == ListCount. O(n) is the time complexity.

ListCount is the number of elements in the list. Divide ListCount by 2. Keep start pointer in head. set another startCount variable to 0. Increment startCount by one and reach the next node till start count == ListCount. O(n) is the time complexity.

hair and tortoise algorithm

a = ll.next ;
b = ll.next.next ;

while (b != null)
{ b=b.next.next;
a =a.next; }

return a;

PS :- there is nothing like middle node for ever number of node .. works only if number of nodes is more than 1.