设计一个支持下述操作的食物评分系统:
修改 系统中列出的某种食物的评分。
返回系统中某一类烹饪方式下评分最高的食物。
实现 FoodRatings
类:
FoodRatings(String[] foods, String[] cuisines, int[] ratings)
初始化系统。食物由foods
、cuisines
和ratings
描述,长度均为n
。foods[i]
是第i
种食物的名字。cuisines[i]
是第i
种食物的烹饪方式。ratings[i]
是第i
种食物的最初评分。
void changeRating(String food, int newRating)
修改名字为food
的食物的评分。String highestRated(String cuisine)
返回指定烹饪方式cuisine
下评分最高的食物的名字。如果存在并列,返回 字典序较小 的名字。
注意,字符串 x
的字典序比字符串 y
更小的前提是:x
在字典中出现的位置在 y
之前,也就是说,要么 x
是 y
的前缀,或者在满足 x[i] != y[i]
的第一个位置 i
处,x[i]
在字母表中出现的位置在 y[i]
之前。
示例:
输入
["FoodRatings", "highestRated", "highestRated", "changeRating", "highestRated", "changeRating", "highestRated"]
[[["kimchi", "miso", "sushi", "moussaka", "ramen", "bulgogi"], ["korean", "japanese", "japanese", "greek", "japanese", "korean"], [9, 12, 8, 15, 14, 7]], ["korean"], ["japanese"], ["sushi", 16], ["japanese"], ["ramen", 16], ["japanese"]]
输出
[null, "kimchi", "ramen", null, "sushi", null, "ramen"]
解释
FoodRatings foodRatings = new FoodRatings(["kimchi", "miso", "sushi", "moussaka", "ramen", "bulgogi"], ["korean", "japanese", "japanese", "greek", "japanese", "korean"], [9, 12, 8, 15, 14, 7]);
foodRatings.highestRated("korean"); // 返回 "kimchi"
// "kimchi" 是分数最高的韩式料理,评分为 9 。
foodRatings.highestRated("japanese"); // 返回 "ramen"
// "ramen" 是分数最高的日式料理,评分为 14 。
foodRatings.changeRating("sushi", 16); // "sushi" 现在评分变更为 16 。
foodRatings.highestRated("japanese"); // 返回 "sushi"
// "sushi" 是分数最高的日式料理,评分为 16 。
foodRatings.changeRating("ramen", 16); // "ramen" 现在评分变更为 16 。
foodRatings.highestRated("japanese"); // 返回 "ramen"
// "sushi" 和 "ramen" 的评分都是 16 。
// 但是,"ramen" 的字典序比 "sushi" 更小。
提示:
1 <= n <= 2 * 104
n == foods.length == cuisines.length == ratings.length
1 <= foods[i].length, cuisines[i].length <= 10
foods[i]
、cuisines[i]
由小写英文字母组成1 <= ratings[i] <= 108
foods
中的所有字符串 互不相同在对
changeRating
的所有调用中,food
是系统中食物的名字。在对
highestRated
的所有调用中,cuisine
是系统中 至少一种 食物的烹饪方式。最多调用
changeRating
和highestRated
总计2 * 104
次
题解:
class FoodRatings:
def __init__(self, foods: List[str], cuisines: List[str], ratings: List[int]):
self.food_map = {} # 食物为键的字典
self.cuisine_map = defaultdict(SortedList) # 烹饪方式为键的字典
for food, cuisine, rating in zip(foods, cuisines, ratings):
self.food_map[food] = [rating, cuisine]
self.cuisine_map[cuisine].add((-rating, food)) #从大到小排序,字典序小到大排序
def changeRating(self, food: str, newRating: int) -> None:
rating, cuisine = self.food_map[food]
sl = self.cuisine_map[cuisine]
sl.discard((-rating, food)) # 删除旧数据
sl.add((-newRating, food)) # 添加新数据
self.food_map[food][0] = newRating #更改fodd对应评分
def highestRated(self, cuisine: str) -> str:
return self.cuisine_map[cuisine][0][1]
# Your FoodRatings object will be instantiated and called as such:
# obj = FoodRatings(foods, cuisines, ratings)
# obj.changeRating(food,newRating)
# param_2 = obj.highestRated(cuisine)