leetcode-1418-点菜展示表
[博客链接]
[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,foodIt
emi] ，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"],["Melis
sa","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。
Related Topics 数组 哈希表 字符串 有序集合 排序
👍 37 👎 0
[题目链接]
[github地址]
[思路介绍]
思路一：穷举
- 感觉思路很简单，更多的是数据结构的使用
public List<List<String>> displayTable(List<List<String>> orders) {
//set去重标题栏
Set<String> tempMenu = new HashSet<>();
Set<String> tempTable = new HashSet<>();
List<List<String>> res = new ArrayList<>();
for (List<String> list: orders
) {
tempMenu.add(list.get(2));
tempTable.add(list.get(1));
}
List<String> tempTitle = new ArrayList<>();
List<String> tempIndex = new ArrayList<>();
tempTitle.addAll(tempMenu);
tempIndex.addAll(tempTable);
Collections.sort(tempTitle);
Collections.sort(tempIndex, new Comparator<String>() {
@Override
public int compare(String o1, String o2) {
return Integer.parseInt(o1) - Integer.parseInt(o2);
}
});
List<String> title = new ArrayList<>();
title.add("Table");
title.addAll(tempTitle);
//记录标题索引
Map<String,Integer> menuMap = new HashMap<>();
for (int i = 0; i < title.size();i++){
menuMap.put(title.get(i),i);
}
res.add(title);
//列桌号
Map<String,Integer> tableMap = new HashMap<>();
int i = 1;
for ( String index: tempIndex
) {
List<String> table = new ArrayList<>();
table.add(index);
tableMap.put(index,i);
i++;
//初始化
tempMenu.forEach(str ->{
table.add("0");
});
res.add(table);
}
//列菜名
for (List<String> list: orders){
int tableIndex = tableMap.get(list.get(1));
int menuName = menuMap.get(list.get(2));
List<String> cnt = res.get(tableIndex);
cnt.set(menuName,String.valueOf(Integer.parseInt(cnt.get(menuName)) + 1));
}
return res;
}
//补充一下测试用例吧
public static void main(String[] args) {
Solution solution = new DisplayTableOfFoodOrdersInARestaurant().new Solution();
List<String> l1 = new ArrayList<>();
List<String> l2 = new ArrayList<>();
List<String> l3 = new ArrayList<>();
List<String> l4 = new ArrayList<>();
List<String> l5 = new ArrayList<>();
List<String> l6 = new ArrayList<>();
List<List<String>> orders = new ArrayList<>();
l1.add("David");
l1.add("3");
l1.add("Ceviche");
l2.add("Corina");
l2.add("10");
l2.add("David");
l3.add("Beef Burrito");
l3.add("3");
l3.add("Fried Chicken");
l4.add("Carla");
l4.add("5");
l4.add("Water");
l5.add("Carla");
l5.add("5");
l5.add("Ceviche");
l6.add("Rous");
l6.add("3");
l6.add("Ceviche");
orders.add(l1);
orders.add(l2);
orders.add(l3);
orders.add(l4);
orders.add(l5);
orders.add(l6);
System.out.println(solution.displayTable(orders));
}
