771. Jewels and Stones

最新推荐文章于 2020-12-27 23:49:46 发布

幸运小猴子

最新推荐文章于 2020-12-27 23:49:46 发布

阅读量82

点赞数

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/sandrew0916/article/details/88432121

本文介绍了一个算法问题，即计算给定字符串中属于特定类型字符的数量。通过C++和Python实现，该问题要求从一系列石头中找出有多少是宝石，区分大小写。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

c++:

class Solution {
public:
    int numJewelsInStones(string J, string S) 
    {
        int count=0;
        set<char> set;
        for(int i=0;i<J.length();i++)
        {
            set.insert(J[i]);
        }
        for(int i=0;i<S.length();i++)        
        {
            if(set.count(S[i]))
                count++;
            
        }
        return count;
    }
};

python:

class Solution(object):
    def numJewelsInStones(self, J, S):
        """
        :type J: str
        :type S: str
        :rtype: int
        """
        j=set(list(J))
        s=list(S)
        count = 0
        for i in s:
            if i in j:
                count += 1
        return count