Leetcode 1418：点菜展示表（超详细的解法！！！）

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。

解题思路

这个问题思路比较简单，但是要将代码组织好不太容易，一个比较容易的思路就是先通过set存储食品和桌号，然后分别按字典序和大小排序。

这个表可以分成三个部分，食品标题，桌号和餐品数量，前面两个通过上述操作已经得到了，对于餐品数量，我这里通过二维字典构造。

class Solution:
    def displayTable(self, orders: List[List[str]]) -> List[List[str]]:
        foods, tables = set(), set()
        for _, i, j in orders:
            tables.add(i)
            foods.add(j)
            
        foods = sorted(list(foods))
        tables = sorted(list(tables))
        res = [["Table"] + foods]
            
        data = {i: {j: 0 for j in foods} for i in tables}
        for _, i, j in orders:
            data[i][j] += 1
        
        for i, j in sorted(data.items(), key=lambda x: int(x[0])):
            res.append([str(i)] + [str(v) for _, v in sorted(j.items())])
        return res

一个更加简洁的写法：

class Solution:
    def displayTable(self, orders: List[List[str]]) -> List[List[str]]:
        tf_cnt = collections.Counter([(T, F) for _, T, F in orders])
        foods  = sorted(set(F for _, _, F in orders))
        tables  = sorted(set(T for _, T, _ in orders), key=int)
        return   [["Table"] + foods] + [[T] + [str(tf_cnt[T, F]) for F in foods] for T in tables]

reference:

https://leetcode.com/problems/display-table-of-food-orders-in-a-restaurant/discuss/586514/Python-2-easyclean-solutions%3A-(A)-4-line-(B)-An-expanded-one-with-explanation

我将该问题的其他语言版本添加到了我的GitHub Leetcode

如有问题，希望大家指出！！！