What pattern does 'Display Table of Food Orders in a Restaurant' follow?

It follows an array scanning plus hash lookup pattern where each table-food pair is counted using a hash map.

How do I ensure the header row is correct?

Collect all unique food items from orders and sort them alphabetically, then prepend 'Table' to form the header.

Should table numbers be treated as strings or integers?

Sort table numbers numerically as integers but convert to strings when constructing the final display table.

How to handle multiple orders of the same food at the same table?

Increment the count in the hash map for that table-food pair each time the item appears.

What are the main failure modes for this problem?

Failing to sort food items alphabetically, miscounting duplicate orders, and not sorting tables numerically are common mistakes.

#1418

Medium

auto_awesome数组·哈希·扫描

LeetCode 题解工作台

点菜展示表

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]=[customerName i ,tableNumber i ,foodItem i ] ，其中 customerName i 是客户的姓名， tableNumber i 是客户所在餐桌的桌号，而 foodIt…

数组哈希表字符串排序有序集合

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]=[customerName_i,tableNumber_i,foodItem_i] ，其中 customerName_i 是客户的姓名，tableNumber_i 是客户所在餐桌的桌号，而 foodItem_i 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 <= customerName_i.length, foodItem_i.length <= 20
customerName_i 和 foodItem_i 由大小写英文字母及空格字符 ' ' 组成。
tableNumber_i 是 1 到 500 范围内的整数。

lightbulb

解题思路

方法一：哈希表 + 排序

我们可以用一个哈希表 $\textit{tables}$ 来存储每张餐桌点的菜品，用一个集合 $\textit{items}$ 来存储所有的菜品。

遍历 $\textit{orders}$ ，将每张餐桌点的菜品存入 $\textit{tables}$ 和 $\textit{items}$ 中。

然后我们将 $\textit{items}$ 排序，得到 $\textit{sortedItems}$ 。

接下来，我们构建答案数组 $\textit{ans}$ ，首先将标题行 $\textit{header}$ 加入 $\textit{ans}$ ，然后遍历排序后的 $\textit{tables}$ ，对于每张餐桌，我们用一个计数器 $\textit{cnt}$ 来统计每种菜品的数量，然后构建一行 $\textit{row}$ ，将其加入 $\textit{ans}$ 。

最后返回 $\textit{ans}$ 。

时间复杂度 $O(n + m \times \log m + k \times \log k + m \times k)$ ，空间复杂度 $O(n + m + k)$ 。其中 $n$ 是数组 $\textit{orders}$ 的长度，而 $m$ 和 $k$ 分别表示菜品种类数和餐桌数。

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class Solution:
    def displayTable(self, orders: List[List[str]]) -> List[List[str]]:
        tables = defaultdict(list)
        items = set()
        for _, table, foodItem in orders:
            tables[int(table)].append(foodItem)
            items.add(foodItem)
        sorted_items = sorted(items)
        ans = [["Table"] + sorted_items]
        for table in sorted(tables):
            cnt = Counter(tables[table])
            row = [str(table)] + [str(cnt[item]) for item in sorted_items]
            ans.append(row)
        return ans

speed

复杂度分析

指标	值
时间	complexity is O(n + m log m + k log k) where n is number of orders, m is number of unique tables, and k is number of unique food items due to scanning, sorting tables, and sorting food items. Space complexity is O(m*k) to store the count mapping of tables to food items.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Candidate correctly identifies array scanning combined with hash map counting.
question_mark
Candidate properly sorts tables numerically and food items alphabetically.
question_mark
Candidate handles edge cases like tables with missing orders for certain food items.

warning

常见陷阱

外企场景

error
Forgetting to sort the food items alphabetically in the header.
error
Mixing up table numbers as strings versus integers when sorting.
error
Incorrectly counting multiple orders of the same food item for a table.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Return a display table showing only the top N most ordered food items.
arrow_right_alt
Include customer names in a secondary structure alongside food counts per table.
arrow_right_alt
Handle streaming orders where counts need to update dynamically.

help

常见问题

外企场景

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