推入并弹出堆栈

Question

This is the push and pop functions written by Codility (source: https://codility.com/media/train/5-Stacks.pdf )

stack = [0] * N
size = 0
def push(x):
    global size
    stack[size] = x
    size = size + 1
def pop():
    global size
    size = size - 1
    return stack[size]

I see that the only thing that the function pop() does is to decrement the value of size by 1, and return stack[size]. How does it actually remove the last element from the stack? I don't really see it...

Answer 1

You need to distinguish between the stack as a concept and the list which implements it. In this implementation, the stack only consists of elements of the list whose index is less than the stack size.

When the stack size is 0, there is nothing on the stack, because there are no list indices less than 0.

After the first push, the stack size is 1, so there is exactly one list element that corresponds to a stack element: stack[0] .

If you then pop an element, the stack size is again reduced from 1 to 0, so the stack is again empty.

At no point does the size of the list change, only the number of list elements that are considered to represent stack elements.

You may be expecting an implementation in which there is always a one-to-one correspondence between stack elements and list elements:

stack = []

pop = stack.pop
push = stack.append

Answer 2

看来，下次调用push（）时，它只是覆盖了元素。