You‘re given strings J representing the types of stones that are jewels, and S representing the stones you have. Each character in Sis 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
题意：给出两个字符串，一个是要匹配的模式串，一个是要匹配的字符串，模式串每个字母都是要匹配的字母，统计目标串中模式串各个字母出现的次数。
想法：最粗暴的想法就是去把模式串中的字母存成像Map或者List的结构数据，然后用的时候去依次扫，如果考虑找是O(1)的话就直接是O（N），（模式串长M，目标串N），如果找是O（M），就是O（MN），目测map的查找应该是O（$\sqrt(M)$）.后面我的做法是先分别排序，然后合并查找，利用的是两者依然有序的特点。具体可以参考下下面的代码，我看了用一行的做法，是利用了python本身map的用法，map(func, iterable)，把func依次作用到iterable的元素上，值得注意的是list.count(str)这个函数，是去匹配list里面出现了多少个str。
我的代码实现：
class Solution(object):
def numJewelsInStones(self, J, S):
"""
:type J: str
:type S: str
:rtype: int
"""
target = sorted(J)
source = sorted(S)
i = len(target)
j = len(source)
index1 = 0
index2 = 0
cnt = 0
while index1 < j and index2 < i:
if source[index1] < target[index2]:
index1+=1
elif source[index1] == target[index2]:
index1+=1
cnt+=1
else:
index2+=1
return cnt
一行的实现：
def numJewelsInStones(self, J, S):
return sum(map(J.count, S))