LeetCode From Easy To Hard No2[Easy]: Jewels and Stones珠宝和石头

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.

普通解法：

class Solution {
    public int numJewelsInStones(String J, String S) {
        int count = 0;
        String[] s = S.split("");
        for (int i = 0; i < s.length; i++) {
            if (J.contains(s[i])) {
                count++;
            }
        }
        return count;
    }
}

利用哈希优化时间复杂度, HashSet 的contain方法是O(1)而String的contain方法是O(n):

class Solution {
    public int numJewelsInStones(String J, String S) {
        Set<Character> setJ = new HashSet<Character>();
            int count = 0;
            for (int i = 0; i < J.length(); i++) {
                setJ.add(J.charAt(i));
                
            }
            for (int i = 0; i < S.length(); i++) {
                if (setJ.contains(S.charAt(i))) {
                    count++;
                }
            }
            return count;
    }
}

 

参考别人的骚操作还有正则表达式一行代码解法，不过并没有用hashset效率高：

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