Passo 22

Instruções

Em seguida, crie um bloco else que atribui current_node.next a self.head. Isso trata o caso em que o elemento a ser removido é encontrado no nó head. Ao atualizar self.head para apontar para current_node.next, o nó head é removido da lista ligada.

O que fazer:

Testes:

Você deve criar um bloco `else` que atribui `current_node.next` a `self.head`.

class LinkedList:
    class Node:
        def __init__(self, element):
            self.element = element
            self.next = None
            
    def __init__(self):
        self.length = 0
        self.head = None

    def is_empty(self):
        return self.length == 0
    
    def add(self, element):
        node = self.Node(element)
        if self.is_empty():
            self.head = node
        else:
            current_node = self.head
            while current_node.next is not None:
                current_node = current_node.next
            current_node.next = node
        self.length += 1

    def remove(self, element):
        previous_node = None
        current_node = self.head
        while current_node is not None and current_node.element != element:
            previous_node = current_node
            current_node = current_node.next

        if current_node is None:
            return        
        elif previous_node is not None:
            previous_node.next = current_node.next


my_list = LinkedList()
print(my_list.is_empty())

my_list.add(1)
my_list.add(2)
print(my_list.is_empty())
print(my_list.length)

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