题目链接：
思路：
从题目得知，我们是求字符串J在字符串S中出现的次数。也就是说，one-pass就可以brute force获得答案。当然可以利用set()数据结构进行优化。算法复杂度：
时间：O(M*N) or O(M + N) where M is the length of J and N is the length of S空间：O(1) or O(M) where M is the length of J
代码：
class Solution(object): def numJewelsInStones(self, J, S): """ :type J: str :type S: str :rtype: int """ count = 0 for s in S: if s in J: count +=1 return count
用set()优化：
class Solution(object): def numJewelsInStones(self, J, S): """ :type J: str :type S: str :rtype: int """ setJ = set(J) return sum(s in setJ for s in S)