1. 题目
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。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/display-table-of-food-orders-in-a-restaurant
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
2. 解题
class Solution {
public:
vector<vector<string>> displayTable(vector<vector<string>>& orders) {
unordered_map<string,map<string,int>> m;//桌号，菜，数量
vector<vector<string>> food = {{"Table"}};//表头
vector<string> tableId;//桌号
for(auto& od : orders)
{
m[od[1]][od[2]]++;
tableId.push_back(od[1]);//桌号
food[0].push_back(od[2]);//菜品
}
sort(tableId.begin(), tableId.end(),[&](auto a, auto b){
if(a.size()==b.size())
return a < b;
return a.size() < b.size();
});//桌号排序
tableId.erase(unique(tableId.begin(),tableId.end()), tableId.end());
//桌号去重
sort(food[0].begin()+1, food[0].end());
//菜品排序
food[0].erase(unique(food[0].begin()+1, food[0].end()), food[0].end());
//菜品去重
food.resize(tableId.size()+1);//开辟空间
for(int i = 1; i < food.size(); ++i)
{
food[i].push_back(tableId[i-1]);//第一列，桌号
for(int j = 1; j < food[0].size(); ++j)
food[i].push_back(to_string(m[food[i][0]][food[0][j]]));
//其余列数量
}
return food;
}
};
984 ms 90.2 MB
