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 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:
S and J will consist of letters and have length at most 50.
The characters in J are distinct.
常规思路
题目可能有点绕，其实简化一下，就是说有两个字符串J和S，两个字符串都是由字母构成的，并且区分大小写，而J里的字母是不重复的，S则是有可能重复的。请找出在S里有多少个字母是与J的字母相同的。
思路很简单，就是遍历S里的每一个字符是否存在于J中，每存在一个就累计一次。代码如下：
class Solution {
public int numJewelsInStones(String J, String S) {
int count = 0;
for (int i = 0; i < S.length(); i++) {
if (J.indexOf(S.charAt(i)) >= 0)
count++;
}
return count;
}
}另辟蹊径
在讨论区里见到了一个很骚的做法，只有一行代码就可以得到想要的结果，如下：
class Solution {
public int numJewelsInStones(String J, String S) {
return S.replaceAll("[^" + J + "]", "").length();
}
}这种思路是利用了正则表达式，把S里凡是不属于J的字符全部替换成空字符串""，然后剩下的字符串的长度就是最终的答案。
至于这两种思路哪个更好就见仁见智了，下面是两个思路各自花费的时间和内存：
常规思路：
Runtime: 1 ms
Memory Usage: 33.7 MB
特殊思路：
Runtime: 7 ms
Memory Usage: 34.9 MB
相关链接