Discussion on: Implement a Stack with TypeScript

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Replies for: Hi Nik, Good question. Yes, that's right you're pushing a value with type string or number but as you can see at right here. push(val: T) { c...

I agree. Let me rephrase. The user of stack should be only aware of stack and the valueType it pushes or pops. It adds no value (in my opinion ) to give back stacknode of popped value.

So, it makes sense to use stacknode internally in stack to keep track of next value but don’t expose stacknode to the user of stack.

Mark Abeto • Jul 20 '20

Awh ok. Sorry for the late reply. I think I understand what you're talking about. You want just the value correct not the object?. If that's what you want, you need two change two parts.

Stack Interface

interface Stack<T> {
  size: number
  top: StackNode<T> | null
  bottom: StackNode<T> | null
  push(val: T): number
  pop(): T | null
}

Stack pop implementation

pop(): T | null {
    if (this.size > 0) {
      const nodeToBeRemove = this.top as StackNode<T>
      this.top = nodeToBeRemove.next
      this.size -= 1
      nodeToBeRemove.next = null
      return nodeToBeRemove.value
    }
    return null
  }