我的任务是使用堆栈评估一个完全带括号的中缀表达式。已为我编写了 Stack 类，我不能更改或修改 Stack 类。
以下是如何评估中缀表达式的分步说明：
只需从左到右扫描表达式。如果它不是a )，则将其压入堆栈。当您遇到 a ) 时，从堆栈中弹出 4 次，进行数学运算并将值压入堆栈。最后，您将在堆栈中只有一个值，这就是答案。
这是该代码：
class Stack:
def __init__(self):
self.theStack=[]
def top(self):
if self.isEmpty():
return "Empty Stack"
else:
return self.theStack[-1]
def isEmpty(self):
return len(self.theStack)==0
def push(self,item):
self.theStack.append(item)
def pop(self):
if not self.isEmpty():
temp=self.theStack[-1]
del(self.theStack[-1])
return temp
else:
return "Empty Stack"
到目前为止，这是我的代码：
def evaluateInfix(Input):
xStack=Stack()
for n in Input:
if n!=")":
print "Pushing %s into the stack" %n
xStack.push(n)
if n==")":
math=xStack.pop()+xStack.pop()+xStack.pop()
last=xStack.pop()
for j in math:
print " Popping %s from stack" %j
print " Popping %s from stack" %last
evaluation=eval(math)
xStack.push(evaluation)
print "Pushing %d into stack" %evaluation
这是我的代码运行的示例：
Enter a fully parenthesized expression that has non-negative integer operands and using only + - * and ( )
Please enter the expression: ((9+9)+(9+9))
Pushing ( into the stack
Pushing ( into the stack
Pushing 9 into the stack
Pushing + into the stack
Pushing 9 into the stack
Popping 9 from stack
Popping + from stack
Popping 9 from stack
Popping ( from stack
Pushing 18 into stack
Pushing + into the stack
Pushing ( into the stack
Pushing 9 into the stack
Pushing + into the stack
Pushing 9 into the stack
Popping 9 from stack
Popping + from stack
Popping 9 from stack
Popping ( from stack
Pushing 18 into stack
Traceback (most recent call last):
File "project2.py", line 252, in <module>
main()
File "project2.py", line 246, in main
Infix=evaluateInfix(Input)
File "project2.py", line 164, in evaluateInfix
math=xStack.pop()+xStack.pop()+xStack.pop()
TypeError: unsupported operand type(s) for +: 'int' and 'str'