LeetCode 771. Jewels and Stones

原题链接在这里：https://leetcode.com/problems/jewels-and-stones/

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.

题解：

Check if there is any char in S appeared in J.

Time Complexity: O(m + n). m = J.length(). n = S.length().

Space: O(m).

AC Java:

 1 class Solution {
 2     public int numJewelsInStones(String J, String S) {
 3         if(J == null || S == null){
 4             return 0;
 5         }
 6         
 7         int count = 0;
 8         HashSet<Character> hs = new HashSet<>();
 9         for(char c : J.toCharArray()){
10             hs.add(c);
11         }
12         
13         for(int i = 0; i < S.length(); i++){
14             if(hs.contains(S.charAt(i))){
15                 count++;
16             }
17         }
18         
19         return count;
20     }
21 }

LeetCode 771. Jewels and Stones

原文：https://www.cnblogs.com/Dylan-Java-NYC/p/12151467.html