[Java-알고리즘] Jewels and Stones

[Problem]

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".

 

[Example1]

Input: J = "aA", S = "aAAbbbb"
Output: 3

 

[Example2]

Input: J = "z", S = "ZZ"
Output: 0

 

[Solution]

class Solution {
    public int numJewelsInStones(String J, String S) {
        int numOfStone = S.length();
        int numOfJewels = J.length();
        
        char[] jewels = J.toCharArray();

        char[] stones = S.toCharArray();
        int numOfMatchedStone = 0;

        
        for(int i = 0; i< numOfStone; i++){
            for(int j = 0; j < numOfJewels; j++){
                if(stones[i] == jewels[j]){
                    numOfMatchedStone++;
                }
            }
        }
    
        return numOfMatchedStone;
    }
}

 

 

 

주어진 돌과 보석 배열을 입력으로 받고, 주어진 돌 중에서 보석이 몇개 있는지 카운트하는 단순한 문제이다.

가장 단순하게 이중 포문을 통해 풀었다.