题目描述

过于水，用来练基础数据结构好啦

样例

def numJewelsInStones(self, J: str, S: str) -> int:
        #可以进一步用set: (O(1)的查询,是个value为0的hashmap)
        # S-> jewelsSet = set(J) 然后查return sum(s in jewelsSet for s in S)
        return sum(J.count(i) for i in S)
        """
        res = 0
        for i in S:
            for j in range(len(J)):
                if i == J[j]:
                    res += 1
        return res
        """

c++暴力

懒人版直接开int

int numJewelsInStones(string J, string S) {
        int res = 0;
        for (char j : J)
            for (char s : S)
                if (j == s)
                    res += 1;
        return res;
    }

或者

int numJewelsInStones(string J, string S) {
        int res = 0;
        for (int i = 0; i < J.size(); i ++) 
            for (int j = 0; j < S.size(); j++)
                if (J[i] == S[j])
                    res += 1;
        return res;
    }