[Leetcode] 771. Jewels and Stones 解题报告

最新推荐文章于 2021-11-13 12:47:40 发布

原创最新推荐文章于 2021-11-13 12:47:40 发布 · 306 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#Leetcode #解题报告 #Hash table

IT公司面试习题专栏收录该内容

736 篇文章

订阅专栏

本文介绍了一个简单的编程问题，即计算给定字符串中属于特定字符集的字符数量。使用哈希表来快速查找并统计匹配字符的数量，展示了如何通过C++实现这一功能。

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 Sis 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.

思路：

练手题目。不过本题除了用哈希表之外，还可以用二分查找树以及向量等数据结构解决，J中的数量不同，将会导致不同的数据结构具有不同的运行效率。

代码：

class Solution {
public:
    int numJewelsInStones(string J, string S) {
        unordered_set<char> hash;
        for (auto j : J) {
            hash.insert(j);
        }
        int ans = 0;
        for (auto s : S) {
            ans += hash.count(s);
        }
        return ans;
    }
};