## Task

Complete function `splitOddAndEven`, accept a number `n`(n>0), return an array that contains the continuous parts of odd or even digits.

Please don't worry about digit `0`, it won't appear ;-)

## Examples

Difficulty: introductory

Starter code:

def split_odd_and_even(n):
	

Solutions:
def split_odd_and_even(n):
    
    import re
  
    return [int(i) for i in re.findall(r"[2468]+|[13579]+", str(n))]

from itertools import groupby

def split_odd_and_even(n):
    return [int("".join(g))
        for i, g in groupby(str(n), key=lambda x: int(x) % 2)]

import re
def split_odd_and_even(n):
    return [int(x) for x in re.findall(r"([2468]+|[13579]+)",str(n))]

def split_odd_and_even(n):
    new_n = ''
    for c in str(n):
        if not new_n or not (int(new_n[-1]) - int(c)) % 2:
            new_n += c
        else :
            new_n += '.'+c
    return [int(i) for i in new_n.split('.')]

def is_even(n):
    return int(n)%2==0

def split_odd_and_even(n):
    index, s = 0, str(n)
    l = [[s[0]]]
    for x in s[1:]:
        if is_even(x) and is_even(l[index][-1]):
            l[index].append(x)
        elif not is_even(x) and not is_even(l[index][-1]):
            l[index].append(x)
        else:
            l.append([x])
            index += 1
    return [int(''.join(x)) for x in l]

from itertools import groupby

def split_odd_and_even(n):
  return [int(''.join(list(gp))) for _, gp in groupby(str(n), key = lambda x: int(x) % 2)]

from itertools import groupby

def split_odd_and_even(n):
    return [int(''.join(grp)) for _, grp in groupby(str(n), key=lambda x: int(x) % 2)]

import re

def split_odd_and_even(n):
    return [int(group[0] or group[1]) for group in re.findall(r'([13579]+)|([2468]+)', str(n))]

from itertools import groupby


def split_odd_and_even(n):
    return [int(''.join(str(b) for b in g)) for k, g in
            groupby((int(a) for a in str(n)), key=lambda b: b % 2)]