[leetcode] 771. Jewels and Stones

原题：

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：

　　用简单的嵌套循环即可实现。

c++代码实现：

class Solution {
public:
    int numJewelsInStones(string J, string S) {
        int num=0;
        for(int i=0;i<J.size();i++)
        {
            for(int j=0;j<S.size();j++)
            {
                if(J[i]==S[j])
                    num++;
            }
        }
        return num;
    }
};

python代码实现：

class Solution:
    def numJewelsInStones(self, J, S):
        """
        :type J: str
        :type S: str
        :rtype: int
        """
        return sum(S.count(j) for j in J)