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 customerName_i 是客户的姓名，tableNumber_i 是客户所在餐桌的桌号，而 foodItem_i 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"]]

限制

• 1 <= orders.length <= 5 * 10^4
• orders[i].length == 3
• 1 <= customerNamei.length, foodItemi.length <= 20
• customerName_i 和 foodItem_i 由大小写英文字母及空格字符 ' ' 组成。
• tableNumber_i 是 1 到 500 范围内的整数。

算法

(模拟) $O(n \log n)$

1. 首先将不同种类的 fooditem 统计出来，并排好序。这里为了方便直接用有序集合。
2. 再遍历 orders，构造表格 res。
3. 根据 Table 编号排序 res 数组。
4. 将 res 数组加上标题，转换成字符串输出。

时间复杂度

空间复杂度

C++ 代码

class Solution {
public:
    vector<vector<string>> displayTable(vector<vector<string>>& orders) {
        set<string> foods;
        for (const auto &v : orders)
            foods.insert(v[2]);

        int cnt = 1;
        unordered_map<string, int> fp;

        for (const auto &s : foods)
            fp[s] = cnt++;


        vector<vector<int>> res;

        cnt = 0;
        unordered_map<string, int> tp;
        for (const auto &v : orders) {
            if (tp.find(v[1]) == tp.end()) {
                tp[v[1]] = cnt;
                res.push_back(vector<int>(foods.size() + 1));
                res[cnt][0] = stoi(v[1]);
                cnt++;
            }

            res[tp[v[1]]][fp[v[2]]]++;
        }

        sort(res.begin(), res.end());

        vector<vector<string>> ans;

        ans.push_back(vector<string>(1));
        ans[0][0] = "Table";

        for (const auto &v : foods)
            ans[0].push_back(v);

        for (int i = 0; i < res.size(); i++) {
            ans.push_back(vector<string>(foods.size() + 1));
            for (int j = 0; j <= foods.size(); j++)
                ans[i + 1][j] = to_string(res[i][j]);
        }

        return ans;
    }
};