【题目】

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".

两个集合A、B。A中是石头，B中是宝石。通过比对返回石头A中是宝石的个数。

【思路】

string转换外char数组，把原始数据存在HashSet中，通过比较键值，是否contains了B中的元素来控制ans++

【代码】

class Solution {

    public int numJewelsInStones(String J, String S) {

        char[] js=J.toCharArray();

        char[] ss=S.toCharArray();

        Set data=new HashSet();

        int ans=;

        //加入数据

        for(char j:js){

            data.add(j);}

        //遍历是否相等

        for(char s:ss){
if(data.contains(s))

                ans++;

            }

        return ans;

    }

}

【参考】