LeetCode 771 - Jewels and Stones

https://leetcode.com/problems/jewels-and-stones/description/

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.

https://leetcode.com/problems/jewels-and-stones/discuss/113553/C++JavaPython-Easy-and-Concise-Solution-O(M+N)

public int numJewelsInStones(String J, String S) {
        int res = 0;
        Set setJ = new HashSet();
        for (char j: J.toCharArray()) setJ.add(j);
        for (char s: S.toCharArray()) if (setJ.contains(s)) res++;
        return res;
    }

https://leetcode.com/problems/jewels-and-stones/discuss/113574/1-liners-PythonJavaRuby

public int numJewelsInStones(String J, String S) {
    return S.replaceAll("[^" + J + "]", "").length();
}

https://leetcode.com/problems/jewels-and-stones/discuss/113562/Java-solution

int res = 0, index = 0;
for (int i = 0; i < J.length(); i++) {
        char jewel = J.charAt(i);
        while ((index = S.indexOf(String.valueOf(jewel), index)) != -1) {
            ++res;
            ++index;
        }
    }
return res;

https://www.cnblogs.com/grandyang/p/8910994.html

这道题给了我们两个字符串，珠宝字符串J和石头字符串S，其中J中的每个字符都是珠宝，S中的每个字符都是石头，问我们S中有多少个珠宝。这道题没什么难度，高于八成的Accept率也应证了其Easy难度实至名归。那么先来暴力搜索吧，就将S中的每个字符都在J中搜索一遍，搜索到了就break掉，