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。
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。
例子
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"]]
方法1.哈希法
核心按照 [[座号1: [菜品1:数量, 菜品2:数量 ...]], [座号2: [菜品1:数量, 菜品2:数量 ...]] ...]
构成哈希数组, 之后再循环按要求构成结果数组
例如:
orders = [["David","3","Ceviche"],["Corina","10","Beef Burrito"],["David","3","Fried Chicken"],["Carla","5","Water"],["Carla","5","Ceviche"],["Rous","3","Ceviche"]]
创建foodArr存放所有食物, tableArr存放座号, temp存放hash数组, 相同菜品数量对应+1
-
第一次遍历
orders-
foodArr= [Ceviche, Beef Burrito, Fried Chicken, Water] -
tableArr= [3, 10, 3, 5, 5, 3] -
temp= [
3:[Ceviche:2, Fried Chicken:1],
10:[Beef Burrito:1],
5:[Water:1, Carla :1]]
-
-
将
foodArr,tableArr按要求正序排列 -
第二次遍历
tableArr, 按照对应key-value对应插入即可
代码
func displayTable(_ orders: [[String]]) -> [[String]] {
var foodArr = [String](), tableArr = [String](),
temp = [String:[String:Int]](), res = [[String]]()
for i in orders {
if !tableArr.contains(i[1]) {
tableArr.append(i[1])
temp[i[1]] = [i[1]:0]
}
if !foodArr.contains(i[2]) {
foodArr.append(i[2])
}
if !temp[i[1]]!.keys.contains(i[2]) {
temp[i[1]]![i[2]] = 1
}else {
temp[i[1]]![i[2]]! += 1
}
}
var res1 = ["Table"];
for i in foodArr.sorted() {
res1.append(i)
}
res.append(res1)
for i in tableArr.sorted(by: { (n1, n2) -> Bool in return Int(n1)! < Int(n2)! }) {
var arr = [i]
for j in 1..<res1.count {
if !temp[i]!.keys.contains(res1[j]) {
arr.append("0")
}else {
arr.append(String(temp[i]![res1[j]]!))
}
}
res.append(arr)
}
return res
}
不过上面看起来很长, 不利于阅读, 精简下代码有
func displayTable(_ orders: [[String]]) -> [[String]] {
var foodArr = [String](), tableArr = [Int: [String: Int]](), res = [[String]]()
for i in orders {
foodArr.append(i[2])
tableArr[Int(i[1])!, default: [:]][i[2], default: 0] += 1
}
let temp = foodArr.sorted(),
titles = [["Table"] + temp]
res = titles + tableArr.sorted { $0.key < $1.key }.map { t, c in
[String(t)] + temp.map { String(c[$0, default: 0]) }
}
return res
}
题目来源：力扣（LeetCode） 感谢力扣爸爸 :)
IOS 算法合集地址
网友评论