checkio的一些题




Xs and Os Referee:


def checkio(l):
    for i in l:
        if i[0] == i[1] and i[0] == i[2] and i[0] != '.':
            return i[0]
    for i in range(3):
        if l[0][i] == l[1][i] and l[0][i] == l[2][i] and l[0][i] != '.':
            return l[0][i]
    if l[0][0] == l[1][1] and l[0][0] == l[2][2] and l[0][0] != '.':
            return l[0][0]
    if l[0][2] == l[1][1] and l[0][2] == l[2][0] and l[0][2] != '.':
            return l[0][2]
    return "D"








First of Tubis for Binary count:


checkio=lambdax:bin(x).count('1')

Numbers Factory:

def checkio(text):
    l = []
    for i in text:
        if i == '(' or i == '[' or i == '{':
            l.append(i)
        if (len(l) != 0) and (i == ')' and l[-1] == '(' or i == ']' and l[-1] == '[' or i == '}' and l[-1] == '{'):
            l.pop()
        elif i == ')' or i == ']' or i == '}':
            return False
    if len(l) == 0:
        return True
    return False
checkio("(((1+(1+1))))]")





Brackets:

def checkio(text):
    l = []
    for i in text:
        if i == '(' or i == '[' or i == '{':
            l.append(i)
        if (len(l) != 0) and (i == ')' and l[-1] == '(' or i == ']' and l[-1] == '[' or i == '}' and l[-1] == '{'):
            l.pop()
        elif i == ')' or i == ']' or i == '}':
            return False
    if len(l) == 0:
        return True
    return False

The Most Wanted Letter

:

def checkio(text):
    text = text.lower()
    m = {}
    for letter in text:
        if not letter.isalpha():
            continue
        if letter not in m:
            m[letter] = 1
        else:
            m[letter] += 1
    a =  sorted(m.items(), key=lambda m: m[1], reverse=True)
    return a[0][0]

The Hamming Distance

:

checkio=lambda m,n:bin(m^n).count('1')