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LEETCODE 1418. Display Table of Food Orders in a Restaurant 解题思路分析

题目大意:

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]=[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。

如果想查看本题目是哪家公司的面试题,请参考以下免费链接: https://leetcode.jp/problemdetail.php?id=1418

解题思路分析:

题目中的人名实际上是没有用的信息,做题时可以忽略掉。本题重点考察的应该是数据的存储结构。解题时我们使用一个HashMap进行数据存储。key是桌号,value是该桌客人的点单内容,而点单内用同样需要使用到HashMap,key是菜名,value是该菜的点单数量。因此数据结构大概如下:

Map<String, Map<String, Integer>> map = new HashMap<>();

理清了数据结构之后其实本题就没有任何难度了,在输出时,只要注意排序就好了。

实现代码:

public List<List<String>> displayTable(List<List<String>> orders) {
    Map<String, Map<String, Integer>> map = new HashMap<>();
    List<String> itemList=new ArrayList();
    for(List<String> order : orders){
        String tableId = order.get(1);
        String itemName = order.get(2);
        if(!itemList.contains(itemName)) itemList.add(itemName);
        Map<String, Integer> countMap = map.getOrDefault(tableId, new HashMap<>());
        int count=countMap.getOrDefault(itemName,0);
        countMap.put(itemName,count+1);
        map.put(tableId, countMap);
    }
    List<String> tableList=new ArrayList(map.keySet());
    Collections.sort(itemList);
    Collections.sort(tableList, (a,b)->Integer.valueOf(a)-Integer.valueOf(b));
    List<List<String>> res = new ArrayList<>();
    itemList.add(0, "Table");
    res.add(itemList);
    for(String tableId : tableList){
        List<String> row = new ArrayList<>();
        row.add(tableId);
        Map<String, Integer> countMap=map.get(tableId);
        for(int i=1;i<itemList.size();i++){
            row.add(countMap.getOrDefault(itemList.get(i),0)+"");
        }
        res.add(row);
    }
    return res;
}

Runtime: 62 ms, faster than 54.90% of Java online submissions for Display Table of Food Orders in a Restaurant.

Memory Usage: 66.1 MB, less than 100.00% of Java online submissions for Display Table of Food Orders in a Restaurant.

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