Title

the problem link

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.

题目大意：就是统计在另一个字符串里出现的字母的个数，用个set就可以了

Solution

//cpp:
class Solution {
public:
	int numJewelsInStones(string J, string S) {
		unordered_set<char> set1;
		for (int i = 0; i < J.length(); i++) {
			set1.insert(J[i]);
		}
		int count = 0;
		for (int i = 0; i < S.length(); i++) {
			if (set1.find(S[i]) != set1.end()) {
				count++;
			}
		}
		return count;
	}
};