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771. Jewels and Stones - Easy

时间:2019-01-02 17:54:15      阅读:142      评论:0      收藏:0      [点我收藏+]

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.
  • The characters in 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;
    }
}

 

771. Jewels and Stones - Easy

原文:https://www.cnblogs.com/fatttcat/p/10209714.html

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