题目

771. 宝石与石头

题解思路

• 题目本身没啥难度，两个for循环就可以解决
• 但是如果使用set可以将时间复杂度优化到o(n)

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注：之前python写多了，需要注意c++中string类中是char

代码

C++

class Solution {
public:int numJewelsInStones(string jewels, string stones) {unordered_map<char, int> a;for (auto &c : jewels){a[c] = 1;}int ans = 0;for (auto &c : stones){if (a.count(c)){ans++;}}return ans;}
};

Python

class Solution:def numJewelsInStones(self, jewels: str, stones: str) -> int:dic = set()ans = 0for ch in jewels:dic.add(ch)for ch in stones:if ch in dic:ans += 1return ans