771. Jewels and Stones
You're given strings J representing the types of stones that are jewels, and Srepresenting 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:
SandJwill consist of letters and have length at most 50.- The characters in
Jare distinct.
题目描述：大概意思是给定两个字符串 J 和 S ，字符串 J 中的任意一个字符在字符串 S 中出现多少次，输出出现的次数。
题目分析：很简单，将 J 字符串中的每一个字符和 S 中的每一个字符进行匹配，如果能够匹配成功，则统计一次匹配成功。
python 代码：
class Solution(object): def numJewelsInStones(self, J, S): """ :type J: str :type S: str :rtype: int """ J_length = len(J) S_length = len(S) count = 0 for i in range(J_length): for j in range(S_length): if J[i] == S[j]: count = count + 1 return count
C++ 代码：
class Solution { public: int numJewelsInStones(string J, string S) { int J_length = J.length(); int S_length = S.length(); int count = 0; for(int i = 0; i < J_length; i++){ for(int j = 0; j < S_length; j++){ if(J[i] == S[j]){ count++; } } } return count; } };
