1418. Display Table of Food Orders in a Restaurant 点菜展示表
- 作者: 负雪明烛
- id: fuxuemingzhu
- 个人博客:https://fuxuemingzhu.cn/
题目地址:https://leetcode-cn.com/contest/weekly-contest-185/problems/reformat-the-string/
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,Water3 ,0 ,2 ,1 ,05 ,0 ,1 ,0 ,110 ,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 * 10^4orders[i].length == 31 <= customerNamei.length, foodItemi.length <= 20customerNamei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。
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。
字典统计边的次数
Section titled “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欢迎关注负雪明烛的刷题博客,leetcode刷题800多,每道都讲解了详细写法!
2020 年 4 月 19 日 —— 近期比赛太多

评论与交流