- 作者: 負雪明燭
- id: fuxuemingzhu
- 個人博客:http://fuxuemingzhu.cn/
題目地址:https://leetcode-cn.com/contest/weekly-contest-185/problems/reformat-the-string/
題目描述
給你一個數組 orders
,表示客戶在餐廳中完成的訂單,確切地說, orders[i]=[customerNamei,tableNumberi,foodItemi]
,其中 customerNamei
是客戶的姓名,tableNumberi
是客戶所在餐桌的桌號,而 foodItemi
是客戶點的餐品名稱。
請你返回該餐廳的 點菜展示表 。在這張表中,表中第一行爲標題,其第一列爲餐桌桌號 “Table”
,後面每一列都是按字母順序排列的餐品名稱。接下來每一行中的項則表示每張餐桌訂購的相應餐品數量,第一列應當填對應的桌號,後面依次填寫下單的餐品數量。
注意:客戶姓名不是點菜展示表的一部分。此外,表中的數據行應該按餐桌桌號升序排列。
示例 1:
輸入:orders = [["David","3","Ceviche"],["Corina","10","Beef Burrito"],["David","3","Fried Chicken"],["Carla","5","Water"],["Carla","5","Ceviche"],["Rous","3","Ceviche"]]
輸出:[["Table","Beef Burrito","Ceviche","Fried Chicken","Water"],["3","0","2","1","0"],["5","0","1","0","1"],["10","1","0","0","0"]]
解釋:
點菜展示表如下所示:
Table,Beef Burrito,Ceviche,Fried Chicken,Water
3 ,0 ,2 ,1 ,0
5 ,0 ,1 ,0 ,1
10 ,1 ,0 ,0 ,0
對於餐桌 3:David 點了 "Ceviche" 和 "Fried Chicken",而 Rous 點了 "Ceviche"
而餐桌 5:Carla 點了 "Water" 和 "Ceviche"
餐桌 10:Corina 點了 "Beef Burrito"
示例 2:
輸入:orders = [["James","12","Fried Chicken"],["Ratesh","12","Fried Chicken"],["Amadeus","12","Fried Chicken"],["Adam","1","Canadian Waffles"],["Brianna","1","Canadian Waffles"]]
輸出:[["Table","Canadian Waffles","Fried Chicken"],["1","2","0"],["12","0","3"]]
解釋:
對於餐桌 1:Adam 和 Brianna 都點了 "Canadian Waffles"
而餐桌 12:James, Ratesh 和 Amadeus 都點了 "Fried Chicken"
示例 3:
輸入:orders = [["Laura","2","Bean Burrito"],["Jhon","2","Beef Burrito"],["Melissa","2","Soda"]]
輸出:[["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
由大小寫英文字母及空格字符' '
組成。tableNumberi
是 1 到 500 範圍內的整數。
題目大意
給出了 Table 和 food 的一些匹配關係,求每條邊出現的次數,以形成一張表格。
解題方法
字典統計邊的次數
這個題本身不難,但是比較噁心,因爲要返回的結果必須是指定格式的。所以我的代碼寫的賊麻煩。
- 統計 foods 和 tables 分別爲多少,並進行排序。
- 統計每個桌的各個菜的次數
- 把所有的桌的菜按照順序拼接成列表
Python代碼如下:
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 —— 近期比賽太多