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".
给定字符串J代表是珠宝的石头类型，而S代表你拥有的石头.S中的每个字符都是你拥有的一个石头. 你想知道你的石头有多少是珠宝。
J中的字母一定不同，J和S中的字符都是字母。 字母区分大小写，因此"a"和"A"是不同的类型。
Examples
Input: J = "aA", S = "aAAbbbb"
Output: 3
Answer
Using hash table to store all the jewels, and then count the jewels appeared in string S.
Time Complexity
O(n + m) which n refers to length of J and m refers to length of S.
Code
C++
class Solution {
public:
int numJewelsInStones(string J, string S) {
if(J.length() == 0) return 0;
unordered_set<char> jews;
for(int i = 0; i < J.length(); ++i)
jews.insert(J[i]);
int res = 0;
for(int i = 0; i < S.length(); ++i)
{
if(jews.find(S[i]) != jews.end())
res++;
}
return res;
}
};
