프로그래머스 - 즐겨찾기가 가장 많은 식당 정보 출력하기

SELECT R.FOOD_TYPE, R.REST_ID, R.REST_NAME, R.FAVORITES
FROM REST_INFO AS R JOIN(
    SELECT FOOD_TYPE, MAX(FAVORITES) AS MAX_F
    FROM REST_INFO
    GROUP BY FOOD_TYPE
) AS I 
ON R.FOOD_TYPE = I.FOOD_TYPE AND R.FAVORITES = I.MAX_F
ORDER BY R.FOOD_TYPE DESC;