Given a mathematical equation that has `*,+,-,/`, reverse it as follows:

```Haskell
solve("100*b/y") = "y/b*100"
solve("a+b-c/d*30") = "30*d/c-b+a"
```

More examples in test cases. 

Good luck!

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Difficulty: EASY
Starter code: def solve(eq):
	

Solutions:
import re

def solve(eq):
	return ''.join(reversed(re.split('(\\W+)', eq)))


def solve(eq):
	leq = eq.replace('*', ' * ').replace('/', ' / ').replace('+', ' + ').replace('-', ' - ').split(' ')
	out = ''.join(leq[::-1])
	return out


def solve(eq):
	s = eq.replace('+', ' + ').replace('-', ' - ').replace('/', ' / ').replace('*', ' * ')
	return ''.join(s.split()[::-1])


def solve(eq):
	r = []
	n = ''
	for i in eq:
		if i in '*/-+':
			r.append(n)
			r.append(i)
			n = ''
		else:
			n += i
	r.append(n)
	r.reverse()
	return ''.join(r)


import re

def solve(eq):
	return ''.join(re.split('([*+-/])', eq)[::-1])


def solve(eq):
	answer = ''
	temp = ''
	for i in eq[::-1]:
		if i.isalnum():
			temp += i
		else:
			answer += temp[::-1] + i
			temp = ''
	answer += temp[::-1]
	return answer


import re

def solve(eq):
	res = re.findall('([^0-9]+)|(\\d+)', eq)[::-1]
	return ''.join([a[::-1] + b for (a, b) in res])


import re
from itertools import zip_longest

def solve(eq):
	val = re.split('[+-/*]', eq)[::-1]
	symb = re.findall('[+-/*]', eq)[::-1]
	return ''.join((i + j for (i, j) in zip_longest(val, symb, fillvalue='')))


def solve(eq):
	eqnLen = len(eq)
	revEqn = ''
	token = ''
	i = 0
	while i < eqnLen:
		token = eq[i]
		while i + 1 < eqnLen and token.isdigit() and eq[i + 1].isdigit():
			i += 1
			token += eq[i]
		revEqn = token + revEqn
		i += 1
	return revEqn