根據(jù)老師寫了一個(gè)整個(gè)的程序，這個(gè)是申請(qǐng)物理空間的，因?yàn)槲锢砜臻g很大一般不會(huì)不足，符合你的問題，所以可以按需要取多少大小。

不是很規(guī)范，有所不足互相學(xué)習(xí)哈

#include<iostream>
using?namespace?std;
#define?elemtype?char
struct?node?{
elemtype?data;
node?*last;
node?*next;
//新定義的結(jié)點(diǎn)的默認(rèn)構(gòu)造函數(shù)?
node(const?elemtype?data?=?0,node?*last?=?NULL,node?*next?=?NULL)?{
this->data?=?data;
this->last?=?last;
this->next?=?next;
}
~node()?{
data?=?0;
last?=?next?=?NULL;
}
};
class?stack{
private:
node?*base;?//棧底?
node?*top;??//棧頂?
int?size;???//棧的實(shí)際大小?
public:
stack()?{
//申請(qǐng)一個(gè)哨兵結(jié)點(diǎn)作為棧底，哨兵的data沒有意義?
//如果申請(qǐng)物理空間失敗，拋出異常?
if?(!(base?=?new?node()))?{
cout?<<?"內(nèi)存不足!";
//throw?...
}
top?=?base;
size?=?0;
}
~stack()?{
node?*p=base,*q=NULL;
while(p)?{
q?=?p;
p?=?p->next;
delete?q;
}
}
bool?isfull()?{
//嘗試申請(qǐng)一個(gè)node，如果可以申請(qǐng)說明未滿?
node?*p=new?node;
if?(!p)?{
return?true;
}
else?{
delete?p;
return?false;
}
}
bool?isempty()?{
return?!size;
//也可以寫成?return?top?==?base;
}
void?push(const?elemtype?&c)?{
if?(isfull())?{
cout?<<?"滿了";
//throw?..?
}
node?*p?=?new?node(c);
top->next?=?p;
p->last?=?top;
top?=?p;
++size;
}
elemtype?pop()?{
if?(isempty())?{
cout?<<?"空了";
//theow?..
}
elemtype?x?=?top->data;
node?*p?=?top;
top?=?p->last;
top->next?=?NULL;
delete?p;
--size;
return?x;
}
//遍歷函數(shù)，默認(rèn)從棧底到棧頂?
void?traverse(bool?isfrombottom?=?true)?{
node?*p;
if?(isfrombottom)?{
p?=?base->next;
while?(p)?{
cout?<<?p->data;
p?=?p->next;
}
cout<<endl;
}
else?{
p?=?top;
while?(p?!=?base)?{
cout?<<?p->data;
p?=?p->last;
}
cout<<endl;
}
}?
int?lenth()?{
return?size;
}
void?clearStack()?{
node?*p?=?base->next,*q;
while?(p)?{
q?=?p;
p?=?p->next;
delete?q;
}
top?=?base;
size?=?0;
base->next?=?NULL;
}
};
int?main(void)?{
stack?*s?=?new?stack();
if?(s->isempty())
cout<<"棧為空"<<endl;
cout?<<?s->lenth()<<endl;
s->push('h');
s->push('e');
s->push('l');
s->push('l');
s->push('o');
cout?<<?s->lenth()<<endl;
s->traverse();
s->traverse(false);
s->clearStack();
cout?<<?s->lenth()<<endl;
cout?<<endl;
s->push('a');
s->traverse();
s->pop();
s->traverse();
//測(cè)試是不是能二次入棧?
s->push('a');
s->traverse();
s->pop();
s->traverse();
//因?yàn)閮?nèi)存物理空間蠻大的，所以一般棧滿的情況不會(huì)出現(xiàn)?在此不做測(cè)試?
//if?(s->isfull())
//cout<<"棧為滿"<<endl;
s->clearStack();
//最后這句話送給大家
s->push('H');
s->push('a');
s->push('p');
s->push('p');
s->push('y');
s->push('?');
s->push('C');
s->push('h');
s->push('r');
s->push('i');
s->push('s');
s->push('t');
s->push('m');
s->push('a');
s->push('s');
s->push('!');
s->traverse();
delete?s;
return?0;
}

用鏈棧，通過new或malloc動(dòng)態(tài)申請(qǐng)內(nèi)存。但結(jié)點(diǎn)之間不再是數(shù)組下標(biāo)之間的關(guān)系，它們是通過指針串起來的。和鏈表相似。

用new寫