Description：
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.

题意：给定字符串J，里面的每个字符表示一种类型的珠宝；给定字符串S，里面的每个字符可能是珠宝也可能是普通的石头；现要求判断字符串S中有多少的珠宝；

解法：我们可以用哈希表存储字符串J中每种类型的珠宝，然后遍历字符串S，判断是否能在表中找到这种类型的珠宝；

class Solution {
    public int numJewelsInStones(String J, String S) {
        //创建哈希表用于表示珠宝
        Set<Character> jewels = new HashSet<Character>();
        //创建变量保存珠宝的数量
        int cnt = 0;
        //将J中每种类型的珠宝保存在jewels中
        for (int i = 0; i < J.length(); i++) {
            jewels.add(J.charAt(i));
        }
        //判断S中有多少珠宝
        for (int i = 0; i < S.length(); i++) {
            if (jewels.contains(S.charAt(i))) {
                cnt++;
            }
        }
        return