如果用链表实现堆栈，堆栈的大小函数的时间复杂度如何？

Question

At Any point of time if we invoke size function it means to calculate size we need to traverse whole linked list which is O(n). I want to know about the standard template library c++ size function.. The only way to get O(1) is to keep a track of push and pop operations ? is it how it is done in c++ standard library?

Answer 1

You don't necessarily need to traverse a list to know its size. As long as you keep the size as a member variable of the class and correctly increment/decrement it when the size of the list changes, returning the size() is a O(1) operation. The size() member function can be implemented as simply size_t size() const noexcept { return m_size; } size_t size() const noexcept { return m_size; } or similar.

Answer 2

According to the container requirements (C++ 17, 26.2.1 General container requirements) if a container has member function size then its complexity is constant. And the standard container std::list has the member function size.

And the member function of the standard container adapter std::stack is defined like

size_type size() const { return c.size(); }

where c is the underlined container as for example std::list . So it also has the constant complexity.