题解 | 宝石计数

#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param jewels string字符串 
# @param stones string字符串 
# @return int整型
#
class Solution:
    def numJewelsInStones(self , jewels: str, stones: str) -> int:
        # write code here
        
            jewels_set=set(jewels)
            return sum(1 for stone in stones if stone in jewels_set)

in set 复杂度是O(1) ,in str是O(len(str))