771. 宝石与石头

题目链接

https://leetcode-cn.com/problems/jewels-and-stones/

题目原文

中文
给定字符串J 代表石头中宝石的类型，和字符串 S代表你拥有的石头。 S 中每个字符代表了一种你拥有的石头的类型，你想知道你拥有的石头中有多少是宝石。
J 中的字母不重复，J 和 S中的所有字符都是字母。字母区分大小写，因此"a"和"A"是不同类型的石头。
示例 1:
输入: J = “aA”, S = “aAAbbbb”
输出: 3
示例 2:
输入: J = “z”, S = “ZZ”
输出: 0
注意:
S 和 J 最多含有50个字母。
J 中的字符不重复。

English
You’re given strings J representing the types of stones that are jewels, and S representing the stones you have. Each character in S is 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
Note:
S and J will consist of letters and have length at most 50.
The characters in J are distinct.

思路方法

方法一：暴力解决

简单的一个for循环，提取S中的每个元素，与J 中元素进行匹配，存在则+1.

class Solution:
    def numJewelsInStones(self, J: str, S: str) -> int:
        count = 0
        for i in S:
            if i in J:
                count += 1        
        return count   

方法二：利用python内置函数

代码来自讨论区，貌似比方法一快一些。这个更加简洁方便。

class Solution:
    def numJewelsInStones(self, J: str, S: str) -> int:
    	return sum(s in J for s in S)

今天决定从easy开始刷起，尽量在三个月内搞定middle level以内的所有题目。

加油！！！