Reverse a Linked List

def reverseList(self, head: ListNode) -> ListNode:
    current = head
    while (current and current.next):
        next = current.next
        current.next = next.next
        next.next = head
        head = next
    return head

• This is a real basic problem, but it can be tricky. You should have it down cold for any interviews.

• Do more of these basic problems series.

• The real key here is memory management, how many pointers do you need to do the task?

• Also do it iteratively (faulty solution to improve on)

def reverseHelper(self, head):
        if (head.next is not None):
            next = head.next
            before = self.reverseHelper(next)
            before.next = head
            head.next = None
            print("before", before)
        print("head", head)
        return head
    # who returns what?
    def reverseList(self, head: ListNode) -> ListNode:
        tail = self.reverseHelper(head)
        print("tail", tail)
        if ( tail):
            return None
        return tail.next