Question

我正在解决挑战，但是自从过去4个小时以来，现在无法解决问题，我被困住了：

挑战：

> Given a mathematical equation that has *,+,-,/, reverse it as follows:
> 
> solve("100*b/y") = "y/b*100" 
> solve("a+b-c/d*30") = "30*d/c-b+a"

我为解决挑战而编写的代码

def solve(s):
    a,b = s.split('/')
    return (b+"/"+a)

预期输出： 'y / b * 100'

观察到的输出： 'y / 100 * b'

在解决此问题时，请您的帮助：

最好的问候，迪瓦卡（Diwakar）

Answer 1

这里是solve函数的实现，可以逆转方程式。

def solve(equation):
    parts = []
    operand = ""
    for ch in equation:
        if ch in ["+", "-", "*", "/"]:
            parts.append(operand)
            parts.append(ch)
            operand = ""
        else:
            operand += ch
    if operand:
        parts.append(operand)

    return "".join(parts[::-1])

solve函数通过将部分（运算符[“ +”，“-”，“ *”，“ /”]和操作数[数字，变量等]）分隔成一个列表来工作。例如。 “ a + bc / d 30”变为[“ a”，“ +”，“ b”，“-”，“ c”，“ /”，“ d”，“ ”，“ 30 “]。反向并加入列表以获取最终解决方案。

Answer 2

def solve(s):              #Hardest part of this problem is to handle NUMBERS.
    li = [s[0]]            # the first element of s,such as "1"
    for i in range(1,len(s)): # begin to handle the rest of s
        if li[-1].isdigit() and s[i].isdigit():  # if the last element of li is digit and the current element of s is also digit,then they belong to a same NUMBER. 
            li[-1] = li[-1] + s[i]
        else:
            li.append(s[i])
    return "".join(li[::-1])

这将奏效，希望对您有帮助

简单方程逆转

2 个答案: