python3__leecode/1418.点菜展示表

一、刷题内容

原文链接

https://leetcode-cn.com/problems/display-table-of-food-orders-in-a-restaurant/

内容描述

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。

二、解题方法

1.方法一：遍历

class Solution:
    def displayTable(self, orders: List[List[str]]) -> List[List[str]]:
        food_set = set()
        for order in orders:
            if order[2] not in food_set:
                food_set.add(order[2])

        food_lst = list(sorted(food_set))
        food_num = len(food_lst)
        food_dic = {food: i for i, food in enumerate(food_lst)}

        table_dic = {}
        for order in orders:
            table, food = int(order[1]), order[2]
            if table not in table_dic:
                table_dic[table] = [0] * food_num
            table_dic[table][food_dic[food]] += 1

        result = [["Table"] + food_lst]
        for table in sorted(table_dic.keys()):
            result.append([str(table)] + [str(n) for n in table_dic[table]])

        return result

2.方法二

class Solution:
    def displayTable(self, orders: List[List[str]]) -> List[List[str]]:
        d=collections.defaultdict(lambda:collections.defaultdict(lambda:0))
        ts,fs=set(),set()
        for _,table,food in orders:
            d[table][food]+=1
            ts.add(table)
            fs.add(food)
        ts,fs=sorted(ts,key=int),sorted(fs)
        res=[['Table']+fs]
        for i in ts:
            res.append([i]+[str(d[i][j]) for j in fs])
        return res

3.方法三

class Solution:
    def displayTable(self, orders: List[List[str]]) -> List[List[str]]:
        dic = {}
        for order in orders:
            if order[1] in dic:
                dic[order[1]].append(order[2])
            else:
                dic[order[1]] = [order[2]]
        # print(dic)
        lst = [["Table"]]
        dishes = set()
        for v in dic.values():
            for i in v:
                dishes.add(i)
        dishes = list(dishes)
        dishes.sort()
        lst[0].extend(dishes)
        # print(dic)
        for k in sorted(dic, key = lambda x: int(x)):
            tmp = [k]
            for d in dishes:
                tmp.append(str(dic[k].count(d)))
            lst.append(tmp)
        return lst

4.方法三

class Solution:
    def displayTable(self, orders: List[List[str]]) -> List[List[str]]:
        table = [defaultdict(int) for _ in range(501)]
        foods = set()
        for cos, t, food in orders:
            table[int(t)][food] += 1
            foods.add(food)
        foods = sorted(list(foods))
        ans = [["Table"] + foods]
        for i in range(1, 501):
            if table[i]:
                tmp = [str(i)]
                for food in foods:
                    tmp.append(str(table[i][food]))
                ans.append(tmp)
        return ans 

5.方法五（best)

class Solution:
    def displayTable(self, orders: List[List[str]]) -> List[List[str]]:
        dishes = set()
        tables = set()

        for order in orders:
            tables.add(order[1])
            dishes.add((order[2]))

        dishes = list(dishes)
        dishes.sort()

        table_dish = {}
        for table in tables:
            table_dish[table] = [0 for i in range(len(dishes))]

        for order in orders:
            table_dish[order[1]][dishes.index(order[2])] += 1

        table_dish = list(table_dish.items())
        table_dish.sort(key=lambda x:int(x[0]))

        display = []
        first_line = ["Table"]
        for dish in dishes:
            first_line.append(dish)
        display.append(first_line)

        for table, dish_count in table_dish:
            line = [table]
            for count in dish_count:
                line.append(str(count))
            display.append(line)
        return display