771. Jewels and Stones

Link: https://leetcode.com/problems/jewels-and-stones/submissions/

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.

---

題目翻譯:

你得到的字符串 J 代表珠寶的石頭類型，S 代表你擁有的寶石。 S 中的每個角色都是你擁有的一種石頭。你想知道你有多少寶石也是珠寶。
J 中的字母保證不同，J 和 S 中的所有字符都是字母。字母區分大小寫，因此“a”被認為是與“A”不同類型的石頭。

程式思路:

建立一個珠寶石頭類型 hash table 在比對自已擁有的珠寶，就能輕鬆解題。

class Solution {
public:
    int numJewelsInStones(string J, string S) {
        unsigned int ret = 0;
        std::unordered_map<char,int> umap;
        for (unsigned i=0; i<J.length(); ++i)
        {
            umap.insert( pair<char, int>( J.at(i), i));
        }

        for (unsigned i=0; i<S.length(); ++i)
        {
            if(umap.find(S.at(i)) != umap.end())
                ret ++;
        }
        return ret;
    }
};