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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 foodscuisinesratings 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 xy 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 changeRatinghighestRated 总计 2 * 104
通过次数
20.4K
提交次数
48.5K
通过率
42.1%


相关企业

提示 1
The key to solving this problem is to properly store the data using the right data structures.

提示 2
Firstly, a hash table is needed to efficiently map each food item to its cuisine and current rating.

提示 3
In addition, another hash table is needed to map cuisines to foods within each cuisine stored in an ordered set according to their ratings.


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["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"]]
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