1418. Display Table of Food Orders in a Restaurant 点菜展示表
2022年3月7日
- 作者: 负雪明烛
- id: fuxuemingzhu
- 个人博客:http://fuxuemingzhu.cn/
@TOC
题目地址:https://leetcode-cn.com/contest/weekly-contest-185/problems/reformat-the-string/
题目描述
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。
题目大意
给出了 Table 和 food 的一些匹配关系,求每条边出现的次数,以形成一张表格。
解题方法
字典统计边的次数
这个题本身不难,但是比较恶心,因为要返回的结果必须是指定格式的。所以我的代码写的贼麻烦。
- 统计 foods 和 tables 分别为多少,并进行排序。
- 统计每个桌的各个菜的次数
- 把所有的桌的菜按照顺序拼接成列表
Python代码如下:
class Solution:
def displayTable(self, orders: List[List[str]]) -> List[List[str]]:
count = collections.defaultdict(dict)
foods = set()
tables = set()
for order in orders:
foods.add(order[2])
tables.add(int(order[1]))
foods = sorted(list(foods))
cols = ["Table", ] + foods
res = []
res.append(cols)
for order in orders:
if order[2] not in count[order[1]]:
count[order[1]][order[2]] = 0
count[order[1]][order[2]] += 1
for table in sorted(list(tables)):
table = str(table)
tc = count[table]
line = [table,]
for food in foods:
if food not in tc:
line.append("0")
else:
line.append(str(tc[food]))
res.append(line)
return res
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日期
2020 年 4 月 19 日 —— 近期比赛太多