表示堆栈的单链接列表

Question

为什么仅当节点之间的链接指向顶部（较新）到底部（较旧）的方向时，为什么我们不能使用单链列表来表示堆栈？

Answer 1

Because otherwise operations like pop() will take O(n) instead of O(1) and same with the other ones. To represent a stack means accessing the last inserted item is the easiest (less time).

Answer 2

Okay i get it now, removing the head of a singly linked list take O(1) time and removing the tail takes O(n) time since you have to do a linear search to find the node that points to the tail and make that node point to null. So if the stack was tail to head (each pointer points up towards top of stack) it would take O(n) every time you popped as opposed to O(1) for a stack thats implemented head to tail (each pointer points to bottom of stack).