简单的计算程序

今天的题目是:

如何写出一个简单的4则运算的程序。如输入 ” 1 -2*3 + 6/3″ ,输出 -3。只用支持正整数,不用考虑有括号的情况。

思路:因为不用考虑括号,所以乘和除的优先级最高。可以先把输入字符串用“+”和“-”分隔成一个数组,数组的结构是一个数字和符号间隔,把乘除法的项当作一个数字处理,如题目中的输入会被分成[“1″,”-“,”2*3″,”+”,”6/3″],然后再每三个数组项计算一次,把得到的结果加入到剩下的数组里,递归计算。如果计算过程中碰到还有乘除法的项,如”2*3″,则先把这个项算出来。

具体代码如下:

 #!/usr/bin/python  # encoding:UTF-8    '''  Calculate the input string and output the result.  For input of '1-2*3+6/3', output -3  Only consider the situation of int, assume no "(" in the string  '''  add_sub = ['-','+']  mul_div = ['*','/']    def parse(input, pattern):      '''          parse the input string to a raw list          every elemtns in the list is either an operand or operator          treat add/substruct and multiply/division differently          the operand of add/sub could be a expression of mult/div      '''      raw_list = []      digits = ''      index = 0      for c in input.strip():          index += 1          if c == ' ':              #check if space              continue          elif c not in pattern:              #if not a operator store it              digits += c              if index == len(input):                  raw_list.append(digits)          elif c in pattern:              #if find a operator, append the formal found digits              #to the output list              raw_list.append(digits)              raw_list.append(c)              digits = ''      return raw_list    def cal_unit(expression):      oper1, sign, oper2 = expression      oper1 = check_if_mult_div(oper1)      oper2 = check_if_mult_div(oper2)      if sign == '+':          result = oper1+oper2      if sign == '-':          result = oper1-oper2      if sign == '/':          result = oper1*1.0/oper2      if sign == '*':          result = oper1*oper2      return str(result)    def check_if_mult_div(operand):      if operand.find('/') == -1 and operand.find('*') == -1:          return float(operand.strip())      else:          #if the operand contains * or /, calculate it          return float(cal_all(parse(operand, mul_div)))    def cal_all(list):      #user recursion to calculate the whole list      if len(list) == 3:          return cal_unit(list)      else:          return cal_all([cal_unit(list[:3])]+list[3:])    if __name__=='__main__':      a = raw_input('Enter:')      print cal_all(parse(a, add_sub))