<easy>LeetCode Problem -- 771. Jewels and Stones-优快云博客

本文介绍了一种算法，用于计算给定的石头中宝石的数量。通过使用map对石头的种类和数目进行统计，然后检查哪些是宝石。案例包括区分大小写的情况。

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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

分析：给定一个字符串代表宝石，另一个字符串代表石头，要求检测石头中有多少宝石，其实就是检测J字符串字母在S字符串中的出现的数目，需要注意的是检测的是单个字母而不是J整个字符串。
思路一：用map对石头的种类和数目进行统计，之后用J中的字母去查找map就可以了。

class Solution {
public:
    int numJewelsInStones(string J, string S) {
        map<char, int> stone_box;
        int jewels_num = 0;
        for (int i = 0; i < S.size(); i++) {
            if (stone_box.find(S[i]) == stone_box.end()) {
                stone_box[S[i]] = 1;
            } else {
                stone_box[S[i]] += 1;
            }
        }
        for (int i = 0; i < J.size(); i++) {
            if (stone_box.find(J[i]) != stone_box.end()) {
                jewels_num += stone_box.find(J[i]) -> second;
            }
        }
        return jewels_num;
    }
};