There is a sale in a supermarket, there will be a discount
every n
customer.
There are some products in the supermarket where the id of the i-th
product is products[i]
and the price per unit of this product is prices[i]
.
The system will count the number of customers and when the n-th
customer arrive he/she will have a discount
on the bill. (i.e if the cost is x
the new cost is x - (discount * x) / 100
). Then the system will start counting customers again.
The customer orders a certain amount of each product where product[i]
is the id of the i-th
product the customer ordered and amount[i]
is the number of units the customer ordered of that product.
Implement the Cashier
class:
-
Cashier(int n, int discount, int[] products, int[] prices)
Initializes the object with n
, the discount
, the products
and their prices
. -
double getBill(int[] product, int[] amount)
returns the value of the bill and apply the discount if needed. Answers within 10^-5
of the actual value will be accepted as correct.
Example 1:
Input
["Cashier","getBill","getBill","getBill","getBill","getBill","getBill","getBill"]
[[3,50,[1,2,3,4,5,6,7],[100,200,300,400,300,200,100]],[[1,2],[1,2]],[[3,7],[10,10]],[[1,2,3,4,5,6,7],[1,1,1,1,1,1,1]],[[4],[10]],[[7,3],[10,10]],[[7,5,3,1,6,4,2],[10,10,10,9,9,9,7]],[[2,3,5],[5,3,2]]]
Output
[null,500.0,4000.0,800.0,4000.0,4000.0,7350.0,2500.0]
Explanation
Cashier cashier = new Cashier(3,50,[1,2,3,4,5,6,7],[100,200,300,400,300,200,100]);
cashier.getBill([1,2],[1,2]); // return 500.0, bill = 1 * 100 + 2 * 200 = 500.
cashier.getBill([3,7],[10,10]); // return 4000.0
cashier.getBill([1,2,3,4,5,6,7],[1,1,1,1,1,1,1]); // return 800.0, The bill was 1600.0 but as this is the third customer, he has a discount of 50% which means his bill is only 1600 - 1600 * (50 / 100) = 800.
cashier.getBill([4],[10]); // return 4000.0
cashier.getBill([7,3],[10,10]); // return 4000.0
cashier.getBill([7,5,3,1,6,4,2],[10,10,10,9,9,9,7]); // return 7350.0, Bill was 14700.0 but as the system counted three more customers, he will have a 50% discount and the bill becomes 7350.0
cashier.getBill([2,3,5],[5,3,2]); // return 2500.0
Constraints:
-
1 <= n <= 10^4
-
0 <= discount <= 100
-
1 <= products.length <= 200
-
1 <= products[i] <= 200
- There are not repeated elements in the array
products
. -
prices.length == products.length
-
1 <= prices[i] <= 1000
-
1 <= product.length <= products.length
-
product[i]
exists in products
. -
amount.length == product.length
-
1 <= amount[i] <= 1000
- At most
1000
calls will be made to getBill
. - Answers within
10^-5
of the actual value will be accepted as correct.
题目出这么长的leetcode实属人间之屑
题意是设计一个class,能初始化cashier和getbill,每n个人有一个可以获得discount
class Cashier {
public int mod = 0;
int n = 0;
double disc = 0;
private int[] products;
private int[] prices;
public Cashier(int n, int discount, int[] products, int[] prices) {
this.n = n;
disc = discount;
this.products = products;
this.prices = prices;
}
public double getBill(int[] product, int[] amount) {
mod ++;
double bill = 0;
for(int i = 0; i < product.length; i++){
for(int j = 0; j < products.length; j++){
if(products[j] == product[i]) bill += prices[j] * amount[i];
}
}
return mod % n == 0 ? (bill*(1 - disc / 100)) : bill;
}
}
方法一:一时大意忘了hashmap这个东西,就感觉怪怪的
class Cashier {
private int mod = 0;
private Map<Integer, Integer> map;
private double disc = 0;
private int n = 0;
public Cashier(int n, int discount, int[] products, int[] prices) {
this.n = n;
disc = discount;
this.map = new HashMap();
for(int i = 0; i < products.length; i++) map.put(products[i], prices[i]);
}
public double getBill(int[] product, int[] amount) {
mod ++;
double bill = 0;
for(int i = 0; i < product.length; i++){
bill += map.get(product[i]) * amount[i];
}
return mod % n == 0 ? (bill*(1 - disc / 100)) : bill;
}
}
用hashmap来记录products和prices