暴力解法
class Solution:
def numJewelsInStones(self, J: str, S: str) -> int:
total = 0
for j in J:
for s in S:
if s == j:
total+=1
return total
很简单，就不多说了，依次遍历，复杂度O(N²)
hash表
class Solution:
def numJewelsInStones(self, J: str, S: str) -> int:
mp = {x: 0 for x in J}
for s in S:
if mp.get(s) is not None:
mp[s] += 1
return sum(mp.values())
- 先创建一个map, 里面存放类型
- 遍历字符串，如果找到了类型，则map里面的值+1
- 累加所有map的value
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