"""
Implementation of Linked Lists

@jlengrand
2013/12
"""

class SingleListItem():
    """
    Item from a Single Linked List.
    Contains only a previous next element, and a value
    """
    def __init__(self, value, nexti=None):
        self.value = value
        self.nexti = nexti # pointer of the next element in the list

    def has_next(self):
		"""
		Returns True if the item has a next value
		"""
		return not (self.nexti is None)

class SingleLinkedList():
	"""
	Linked list with only one link between items.
	The list can only be traversed one way"""

	def __init__(self):
		self._size = 0
		self._root = None # reference of the head

	def add(self, value):
		"""
		Adds a new element to the end of the list
		"""
		if self._root is None:
			# empty tree
			self._root = SingleListItem(value)
		else:
			curr = self._root
			while(curr.nexti is not None):
				#gets last item
				curr = curr.nexti

			item = SingleListItem(value)
			curr.nexti = item

		self._size += 1

	def delete_item(self, node=None):
		"""
		Deletes the nth node of the list. (0 being the first element)
		If node is None, deletes the first element of the list.
		"""
		if node == None:
			item = self._root.nexti
			self._root = item
			self._size -= 1
		elif node >= self.size():
			raise Exception("Requested value out of list bounds")




	def __len__(self):
		# Returns the number of elements in the list
		return self._size

	def __str__(self):
		"""
		Prints out all values in the list
		This way of doing is not exactly clever, given the fact that I put
		everything in memory before printing
		"""
		to_print = ""
		curr = self._root
		to_print += str(curr.value)
		while(curr.nexti is not None):
			curr = curr.nexti
			to_print += ", " + str(curr.value)

		return to_print

if __name__ == "__main__":
	ll = SingleLinkedList()
	ll.add(3)
	ll.add(4)
	ll.add(5)
	ll.add(6)
	print ll