标签：pes character pac distinct rar different one case ash
You‘re given strings J representing the types of stones that are jewels, and Srepresenting 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.J are distinct.
time: O(M+N), space: O(M) -- M: # of length of J
class Solution { public int numJewelsInStones(String J, String S) { Set<Character> set = new HashSet<>(); for(char c : J.toCharArray()) { set.add(c); } int cnt = 0; for(char c : S.toCharArray()) { if(set.contains(c)) { cnt++; } } return cnt; } }
标签：pes character pac distinct rar different one case ash
原文地址：https://www.cnblogs.com/fatttcat/p/10209714.html
