You're given strings J representing the types of stones that are jewels, and S representing the stones you have. Each character in S is a type of stone you have. You want to know how many of the stones you have are also jewels.
The letters in J are guaranteed distinct, and all characters in J and S are letters. Letters are case sensitive, so "a" is considered a different type of stone from "A".
Example 1:
Input: J = "aA", S = "aAAbbbb"
Output: 3
Example 2:
Input: J = "z", S = "ZZ"
Output: 0
Note:
S and J will consist of letters and have length at most 50.
The characters in J are distinct.
解題思路：
J 用字典或者集合（Python 中判斷集合中元素是否存在也是 O(1) 的時(shí)間）存儲(chǔ)起來，然后遍歷 S 中的每一個(gè)字符，判斷字符是否也出現(xiàn)在 J 中仗岸。
時(shí)間復(fù)雜度 O(M+N)
Python 實(shí)現(xiàn)：
class Solution:
# AC: O(len(J) + len(S))
def numJewelsInStones(self, J, S):
"""
:type J: str
:type S: str
:rtype: int
"""
J = set(J) # Python 中 set 判斷元素是否存在的時(shí)間復(fù)雜度為 O(1)
return sum(ch in J for ch in S)
a = "aA"
b = "aAAbbbb"
print(Solution().numJewelsInStones(a, b)) # 3
