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]=[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。

样例

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" 

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"

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。

算法分析

TreeMap,和TreeSet的运用

• 1、用TreeMap<Integer,HashMap<String,Integer>> food存储(号桌,(菜式，个数))，并按照号桌排序
• 2、用TreeSet<String> set储存菜式，并进行排序
• 3、枚举orders，储存每个桌号的每个菜式的个数分别是多少，最后枚举food中的所有桌号，对于每个桌号，枚举set中所有的菜式，若该桌号有该菜式则输出数量，若没有则输出0

时间复杂度 $O(orders.length * foodItemi.length)$

Java 代码

class Solution {
    public List<List<String>> displayTable(List<List<String>> orders) {
        List<List<String>> ans = new ArrayList();

        TreeMap<Integer,HashMap<String,Integer>> food = new TreeMap<Integer,HashMap<String ,Integer>>();
        Set<String> set = new TreeSet<String>();//所有的食物
        for(List<String> item : orders)
        {
            int id = Integer.parseInt(item.get(1));
            String cai = item.get(2);
            if(!food.containsKey(id)) food.put(id, new HashMap<String,Integer>());
            food.get(id).put(cai, food.get(id).getOrDefault(cai, 0) + 1);
            set.add(cai);
        }

        List<String> one = new ArrayList<String>();
        one.add("Table");
        for(String item : set)
        {
            one.add(item);
        }
        ans.add(one);
        for(int k : food.keySet()) {
            HashMap<String,Integer> t = food.get(k);
            List<String> behind = new ArrayList<String>();
            behind.add(k + "");
            for(String item : set)
            {
                if(t.containsKey(item)) behind.add(t.get(item) + "");
                else behind.add("0");
            }
            ans.add(behind);
        }

        return ans;
    }
}