从堆栈C ++的单个链接列表实现中弹出

Question

I have two functions to pop() in a single-linked list stack. But can't figure out why one is more correct than the other.

POP1:
X = head;
head = head->getNext();
return X;

POP2:
head = head->getNext();
X = head;
return X;

My answer is POP1, but I'm not sure why POP2 would be incorrect. Thanks for any help you can give me.

Answer 1

因为在POP2中，您从堆栈顶部删除了该项目，将其丢弃（即不退回），然后返回了堆栈的当前顶部（实际上是启动时的第二个项目）。

Answer 2

Suppose the list looks like

A -> B -> C -> D -> NULL at this point head=A

So, functionality should remove and return A from the list. Lets see what are POP1 and POP2 doing.

POP1:

X = A;
head = B;
return A; 

At this point list will look like B -> C -> D -> NULL with head pointing to B. And the return value is A. All looks good.

POP2:

head = B;
X = B;
return B; 

At this point list will look like B -> C -> D -> NULL with head also pointing to B. But wait! its returning B instead of A.

Hope this clarifies.