MySQL - 프로그래머스 코딩테스트 [GROUP BY - 즐겨찾기가 가장 많은 식당 정보 출력하기]

SELECT A.FOOD_TYPE, A.REST_ID, A.REST_NAME, A.FAVORITES
  FROM REST_INFO A INNER JOIN (
      SELECT MAX(FAVORITES) AS FAV, FOOD_TYPE
        FROM REST_INFO
       GROUP BY FOOD_TYPE
  ) B ON A.FOOD_TYPE = B.FOOD_TYPE AND A.FAVORITES = B.FAV
 ORDER BY A.FOOD_TYPE DESC;