1418. 点菜展示表
2021-07-06 LeetCode每日一题
链接:https://leetcode-cn.com/problems/display-table-of-food-orders-in-a-restaurant/
标签:数组、哈希表、字符串、有序集合、排序
题目
给你一个数组 orders,表示客户在餐厅中完成的订单,确切地说, orders[i]=[customerNamei,tableNumberi,foodItemi] ,其中 customerNamei 是客户的姓名,tableNumberi 是客户所在餐桌的桌号,而 foodItemi 是客户点的餐品名称。
请你返回该餐厅的 点菜展示表 。在这张表中,表中第一行为标题,其第一列为餐桌桌号 “Table” ,后面每一列都是按字母顺序排列的餐品名称。接下来每一行中的项则表示每张餐桌订购的相应餐品数量,第一列应当填对应的桌号,后面依次填写下单的餐品数量。
注意:客户姓名不是点菜展示表的一部分。此外,表中的数据行应该按餐桌桌号升序排列。
示例 1:
输入:orders = [["David","3","Ceviche"],["Corina","10","Beef Burrito"],["David","3","Fried Chicken"],["Carla","5","Water"],["Carla","5","Ceviche"],["Rous","3","Ceviche"]]
输出:[["Table","Beef Burrito","Ceviche","Fried Chicken","Water"],["3","0","2","1","0"],["5","0","1","0","1"],["10","1","0","0","0"]]
解释:
点菜展示表如下所示:
Table,Beef Burrito,Ceviche,Fried Chicken,Water
3 ,0 ,2 ,1 ,0
5 ,0 ,1 ,0 ,1
10 ,1 ,0 ,0 ,0
对于餐桌 3:David 点了 "Ceviche" 和 "Fried Chicken",而 Rous 点了 "Ceviche"
而餐桌 5:Carla 点了 "Water" 和 "Ceviche"
餐桌 10:Corina 点了 "Beef Burrito"
示例 2:
输入:orders = [["James","12","Fried Chicken"],["Ratesh","12","Fried Chicken"],["Amadeus","12","Fried Chicken"],["Adam","1","Canadian Waffles"],["Brianna","1","Canadian Waffles"]]
输出:[["Table","Canadian Waffles","Fried Chicken"],["1","2","0"],["12","0","3"]]
解释:
对于餐桌 1:Adam 和 Brianna 都点了 "Canadian Waffles"
而餐桌 12:James, Ratesh 和 Amadeus 都点了 "Fried Chicken"
示例 3:
输入:orders = [["Laura","2","Bean Burrito"],["Jhon","2","Beef Burrito"],["Melissa","2","Soda"]]
输出:[["Table","Bean Burrito","Beef Burrito","Soda"],["2","1","1","1"]]
提示:
- 1 <= orders.length <= 5 * 10^4
- orders[i].length == 3
- 1 <= customerNamei.length, foodItemi.length <= 20
- customerNamei 和 foodItemi 由大小写英文字母及空格字符 ' ' 组成。
- tableNumberi 是 1 到 500 范围内的整数。
分析
这题看着就繁琐,但作为调包侠毫不畏惧。
难倒是不难,要理清思路。总结成一句话就是要你求出每道菜在每桌有几个人点了,或者是每一桌点了哪几道菜。所以我要让桌号、菜品名称、数量三个形成联系。这里我的想法是让桌号和菜品名称作为键、然后数量作为值放入哈希表中,这样就能解决数量的问题,另外就是菜品名称要排序,桌号也要排序。
菜品名称和桌号分别放在一个TreeSet中按自然顺序排序,但这里要注意桌号需要转为数字型,如果是字符串,"10"会排在"3"的前面,这就不对了。
根据题目给的条件,菜品数量范围在[1, 20],桌号数量范围在[1, 500],总的订单长度范围是[1, 5 * 10^4],所以按照上面的思路,应该不会超时,但效率应该不咋地。
编码
class Solution {
public List<List<String>> displayTable(List<List<String>> orders) {
List<List<String>> res = new ArrayList<>();
Map<String, Integer> map = new HashMap<>();
// 存放餐品名称,无重复
Set<String> foodItems = new TreeSet<>();
// 存放餐桌,无重复,Integer类型,才能正常排序
Set<Integer> tables = new TreeSet<>();
for (int i = 0; i < orders.size(); i++) {
List<String> list = orders.get(i);
foodItems.add(list.get(2));
tables.add(Integer.parseInt(list.get(1)));
// 桌号 + 餐品名称组成键
String key = list.get(1) + "_" + list.get(2);
map.put(key, map.getOrDefault(key, 0) + 1);
}
// 标题
List<String> head = new ArrayList<>();
head.add("Table");
head.addAll(foodItems);
res.add(head);
for (Integer table : tables) {
List<String> row = new ArrayList<>();
row.add(String.valueOf(table));
for (String foodItem : foodItems) {
String key = table + "_" + foodItem;
Integer value = map.get(key) == null ? 0 : map.get(key);
row.add(String.valueOf(value));
}
res.add(row);
}
return res;
}
}