# # 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 # 返回表达式的值 # @param s string字符串 待计算的表达式 # @return int整型 # class Solution: #判断读入的字符是否为操作符 def is_operation(self,item): if item in {'+','-','*','(',')'}: return True else: return False def cal(self,operand1,operand2,stack_top_opnd): operand1 = int(operand1) operand2 = int(operand2) if stack_top_opnd == '+': res = operand1 + operand2 elif stack_top_opnd == '-': res = operand1 - operand2 elif stack_top_opnd == '*': res = operand1 * operand2 return res #比较操作符的优先级 def operation_compare(self,current_opnd,stack_top_opnd): if stack_top_opnd == "(" and current_opnd == ")": return 0 #括号匹配直接消除 elif stack_top_opnd in {"+","-","*"} and current_opnd == ")": return 1 #计算括号里面的值 elif stack_top_opnd == "("&nbs***bsp;current_opnd == "(": return -1#继续向前走 elif current_opnd == stack_top_opnd: return 1 #可以计算 elif stack_top_opnd =="*": return 1 #也可以计算 elif current_opnd =="*": return -1 #继续向前 else:#最后两者都为加减运算 return 1#也可以计算 def solve(self , s ): # write code here #使用deque定义两个堆栈,分别存放操作符和操作数 from collections import deque operation = deque() operand = deque() current_operand="" s='('+s+')' #加入括号,作为结尾标志 for i in range(len(s)): if self.is_operation(s[i]) is False: #如果为非操作符,则s[i]为操作数部分 current_operand+=s[i] #如果为操作符,则比较当前操作符与operation栈顶操作符的优先级 else: if current_operand!='': operand.append(current_operand) current_operand="" current_opnd = s[i] is_continue = True while(is_continue): if len(operation)==0:#栈顶为空,则将当前操作符入栈,并退出循环 operation.append(current_opnd) is_continue = False else: stack_top_opnd = operation.pop() result = self.operation_compare(current_opnd,stack_top_opnd) if result == 0:#此处考虑左右括号匹配的情况,则当前操作符不入栈,且栈顶元素出栈 is_continue = False elif result == 1:#此处考虑栈顶操作符可运算的情况 operand2 = operand.pop() operand1 = operand.pop() #进行计算 res = self.cal(operand1,operand2,stack_top_opnd) operand.append(res) is_continue = True#计算之后继续比较栈顶元素与当前元素的优先级 elif result == -1:#此处考虑栈顶操作符仍然不可运算的情况,将当前操作符入栈 operation.append(stack_top_opnd) operation.append(current_opnd) is_continue = False return operand.pop()