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
andJ
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:
class Solution { public int numJewelsInStones(String J, String S) { if(J == null || S == null){ return 0; } int count = 0; HashSet<Character> hs = new HashSet<>(); for(char c : J.toCharArray()){ hs.add(c); } for(int i = 0; i < S.length(); i++){ if(hs.contains(S.charAt(i))){ count++; } } return count; } }