每日一题20201105（771. 宝石与石头）

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