771.Jewels and Stones

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字数统计: 256 | 阅读时间 ≈ 1

题目描述

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.

我的解法

解题思路

这题比较简单，使用一个map来记录S中字符出现的次数，然后再在map里查找有没有J中的字符出现。

实现代码

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class Solution {
public:
    int numJewelsInStones(string J, string S) {
        unordered_map<char, int> m;
        for (char c : S)
            m[c]++;
        int res = 0;
        for (char c : J){
            if (m.find(c) != m.end())
                res += m[c];
        }
        return res;
    }
};

执行用时：4 ms, 在所有 C++ 提交中击败了67.30%的用户
内存消耗：6.2 MB, 在所有 C++ 提交中击败了56.14%的用户