LeetCode-771. Jewels and Stones

You're given strings J representing the types of stones that are jewels, and S representing the stones you have.  Each character inS 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".function 1:

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.

function 1:

    First, read J and build jewels hash set.
    Second, read S and count jewels.

    I used hash set so that the time complexity will be O(M+N), instead of O(MN)

class Solution {
    public int numJewelsInStones(String J, String S) {
        int sum=0;
        HashSet<Character> hs = new HashSet<Character>();
        for(char j : J.toCharArray()){
            hs.add(j);
        }
        for(char s : S.toCharArray()){
            if(hs.contains(s))
                sum++;
        }
        return sum;
    }
}

function 2:

Author By StefanPochmann

class Solution {
    public int numJewelsInStones(String J, String S) {
        //利用正则表达式，把S中非J内容去除，replaceAlll()方法,再返回长度
        return S.replaceAll("[^"+J+"]","").length();
    }
}