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本帖最后由 糖逗 于 2020-5-8 14:32 编辑
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。
-  
- 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。
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- class Solution {
- public:
- static bool cmp(string x,string y){
- if(x.size() == y.size())
- return x<y;
- else
- return x.size() < y.size();
- }
-
- vector<vector<string>> displayTable(vector<vector<string>>& orders) {
- set<string> name, order;
- for(int i = 0; i < orders.size(); i++){
- for(int j = 1; j < orders[i].size(); j++){
- if(j == 1) name.insert(orders[i][j]);
- else{
- order.insert(orders[i][j]);
- }
- }
- }
- vector<string> name_(name.begin(), name.end());
- vector<string> order_(order.begin(), order.end());
- sort(order_.begin(), order_.end());
- sort(name_.begin(), name_.end(), cmp);
- int len1 = name_.size(), len2 = order_.size();
- vector<vector<string> > res(len1+1, vector<string>(len2+1, "0"));
- res[0][0] = "Table";
- for(int i = 1; i < res[0].size(); i++) res[0][i] = order_[i-1];
- for(int i = 1; i < res.size(); i++) res[i][0] = name_[i-1];
- for(int i = 0; i < orders.size(); i++){
- auto temp1 = std::find(name_.begin(), name_.end(), orders[i][1]);
- int a = &*temp1 - &name_[0]+1;
- for(int j = 2; j < orders[i].size(); j++){
- auto temp2 = std::find(order_.begin(), order_.end(), orders[i][j]);
- int b = &*temp2 - &order_[0]+1;
- res[a][b] = to_string(stoi(res[a][b])+1);
- }
- }
- return res;
- }
- };
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注意事项:
1、找到vector中对应元素的下标使用std::find 返回的是一个迭代器 再使用 &*temp1 - &name_[0] 得到下标
2、解题思路,先构建行列坐标(TABLE和菜品名称),然后根据行列坐标定位res中单元格进行赋值。
3、- static bool cmp(string x,string y){
- if(x.size() == y.size())
- return x<y;
- else
- return x.size() < y.size();
- }
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注意字符串的排序 |
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