771. Jewels and Stones（python3）

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.

code:

class Solution:
def numJewelsInStones(self, J, S):
"""
:type J: str
:type S: str
:rtype: int
"""
c = 0
for j in J:
c += S.count(j)
return c

tips:
使用python的string的内建函数：str.count(sub, start= 0,end=len(string))
sub – 搜索的子字符串
start – 字符串开始搜索的位置。默认为第一个字符,第一个字符索引值为0。
end – 字符串中结束搜索的位置。字符中第一个字符的索引为 0。默认为字符串的最后一个位置。