Bulls and Cows

注意A了的就不能再放进B里面了

class Solution(object):
    def getHint(self, secret, guess):
        """
        :type secret: str
        :type guess: str
        :rtype: str
        """
        secretDict, guessDict = collections.Counter(), collections.Counter()
        countA, countB = 0, 0

        for index, char in enumerate(secret):
            if char == guess[index]:
                countA += 1
            else:
                secretDict[char] += 1
                guessDict[guess[index]] += 1


        for char in secretDict:
            if char in guessDict:
                countB += min(guessDict[char], secretDict[char])

        return '{}A{}B'.format(countA, countB)

还有一个discussion里的神版本... 主要的是Python里面哈希表还可以用&符号直接实现交集运算......

class Solution(object):
    def getHint(self, secret, guess):
        """
        :type secret: str
        :type guess: str
        :rtype: str
        """
        s, g = collections.Counter(secret), collections.Counter(guess)
        A = sum((1 for i, j in zip(secret, guess) if i == j))
        return '{}A{}B'.format(A, sum((s & g).values()) - A)