利用栈解决括号匹配和逆波兰表达式

当我们学习了栈这种数据结构滞后，我们就可以利用栈来解决一些实际问题。这里是我们给出的动态顺序栈的实现

template< class T>
class Stack
{
public:
Stack()
:_array(new T[3])
, _size(0)
, _capacity(3)
{}
void Push(const T& data)
{
_Checkcapacity();
_array[_size++] = data;

}
void Pop()
{
_size--;
}

bool Empty()
{
if (_size == 0)
return true;
return false;
}
T& Top()
{
return _array[_size-1];
}
T& Top()const
{
return _array[_size - 1];
}
size_t  Size()
{
return _size;
}
size_t Capacity()
{
return _capacity;
}
void  _Checkcapacity()
{
if (_size >= _capacity)
{
T* newarray = new T[_size * 2 + 3];
for (int i = 0; i < _size; i++)
{
newarray[i] = _array[i];
}
_capacity = _size * 2 + 3;
delete _array;
_array = newarray;
}

}
~Stack()
{
if (_array)
delete _array;
_array = NULL;
_size = 0;
_capacity = 0;
}
private:
T * _array;
int _size;
int _capacity;
};

1.括号匹配问题char a[] = "(())abc{[(])}";//左右括号次序不对char b[] = "(()))abc{[]}";//右括号多于左括号char c[] = "(()()abc{[]}";//左括号多于右括号char d[] = "(())abc{[]()}";///左右括号正确bool MatchBrackets(char* c,int size){ Stack<char> ch; for (int i = 0; i < size; i++) {  if (c[i] == '(' || c[i] == '{' || c[i] == '[')  {   ch.Push(c[i]);   continue;  }  else if (c[i] == ')' || c[i] == ']' || c[i] == '}')  {   if (ch.Empty())   {    cout << "右括号多于左括号" << endl;    return true;   }   else   {    if ((c[i]==')'&&ch.Top()=='(')||(c[i] == ']'&&ch.Top() == '[')|| (c[i] == '}'&&ch.Top() == '{'))    {     ch.Pop();     continue;    }    else {     cout << "左右括号匹配次序不正确" << endl;     return false;    }   }  }  else {   continue;  } } if (ch.Empty()) {  cout << "匹配正确" << endl;  return true; } else {  cout << "左括号多于右括号" << endl;  return false; }}2.逆波兰表达式问题实现代码：enum op { OPERAND,OPERATOR, ADD, SUB, MUL, DIV };typedef struct Cell{op _op;int data;}Cell;int RPN(Cell*s, int size){Stack<int> c;for (int i = 0; i < size; i++){if (s[i]._op == OPERAND){c.Push(s[i].data);continue;}else if (s[i]._op == OPERATOR){int right = c.Top();c.Pop();int left = c.Top();c.Pop();switch (s[i].data){case ADD:c.Push(left + right);break;case SUB:c.Push(left - right);break;case MUL:c.Push(left*right);break;case DIV:assert(right != 0);c.Push(left / right);break;default:assert(0);break;}}}return c.Top();}

下面我们给出 表达式4000：

Cell A[] = { {OPERAND,12},{OPERAND,3},{OPERAND,4},{OPERATOR,ADD},{OPERATOR,MUL},{OPERAND,6},{OPERATOR,SUB},{OPERAND,8},{OPERAND,2},{OPERATOR,DIV},{OPERATOR,ADD} };

即 表达式 12*（3+4）-6+8/2  =82；来看程序运行结果：结果正确。