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”.

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.
C/C++解法
最快方法:使用C++的集合，J放入集合中去重，然后遍历S中每一个字符，若在J中则num加1。
`cpp
include
include
include

using namespace std;

// 若要提交到leetcode只需提交class Solution
class Solution {
public:
int numJewelsInStones(string J, string S) {
//set集合所有元素都会根据元素的键值自动排序，set元素的键值就是实值。set不允许两个元素有相同的键值。
set setJ(J.begin(), J.end());
int num = 0;
for (char s : S) {
//.count()记录s在setJ中出现一次,在集合中只可能出现1次或0次
if (setJ.count(s))
num++;
}
return num;
}
};

int main() {
string J_1 = “aA”;
string S_1 = “aAAbbbb”;
string J_2 = “z”;
string S_2 = “ZZ”;
Solution s;
int num_1 = s.numJewelsInStones(J_1, S_1);
int num_2 = s.numJewelsInStones(J_2, S_2);
cout << num_1 << “ “ << num_2 << endl;
return 0;
}
`