简单的方程式反转

Question

我正在解决以下挑战，但自从过去 4 小时以来现在无法解决，我被卡住了：

Challenge :

> 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'

请您帮忙解决这个问题：

此致，迪瓦卡

Answer 1

这里是solve可以反转方程的函数的实现。

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])

求解函数通过将部分（运算符 [“+”、“-”、“*”、“/”] 和操作数 [数字、变量等]）分离到一个列表中来工作。例如。 "a+b-c/d30" 变成 ["a", "+", "b", "-", "c", "/", "d" , " ”，“30”]。反转并加入列表以获得最终解决方案。

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])

这个有用，希望对你有帮助

简单的方程式反转

Simple equation reversal

python

python-3.6