LeetCode Shopping Offers
原题链接在这里:https://leetcode.com/problems/shopping-offers/description/
题目:
In LeetCode Store, there are some kinds of items to sell. Each item has a price.
However, there are some special offers, and a special offer consists of one or more different kinds of items with a sale price.
You are given the each item's price, a set of special offers, and the number we need to buy for each item. The job is to output the lowest price you have to pay for exactly certain items as given, where you could make optimal use of the special offers.
Each special offer is represented in the form of an array, the last number represents the price you need to pay for this special offer, other numbers represents how many specific items you could get if you buy this offer.
You could use any of special offers as many times as you want.
Example 1:
Input: [2,5], [[3,0,5],[1,2,10]], [3,2]
Output: 14
Explanation:
There are two kinds of items, A and B. Their prices are $2 and $5 respectively.
In special offer 1, you can pay $5 for 3A and 0B
In special offer 2, you can pay $10 for 1A and 2B.
You need to buy 3A and 2B, so you may pay $10 for 1A and 2B (special offer #2), and $4 for 2A.
Example 2:
Input: [2,3,4], [[1,1,0,4],[2,2,1,9]], [1,2,1]
Output: 11
Explanation:
The price of A is $2, and $3 for B, $4 for C.
You may pay $4 for 1A and 1B, and $9 for 2A ,2B and 1C.
You need to buy 1A ,2B and 1C, so you may pay $4 for 1A and 1B (special offer #1), and $3 for 1B, $4 for 1C.
You cannot add more items, though only $9 for 2A ,2B and 1C.
Note:
- There are at most 6 kinds of items, 100 special offers.
- For each item, you need to buy at most 6 of them.
- You are not allowed to buy more items than you want, even if that would lower the overall price.
题解:
For recursion, it needs to parameter to calculate current state.
For current state, it needs price and current needs to calcuate direct purchase price.
Then for each offer in special, check if it is valid, if not, skip. If yes, minius offer items in needs and do next level.
This question is bottom up dfs. return the minimum value for current needs.
Time Complexity: O(special.size()*needs.size()*n), n 是recursive call的层数, 可以是needs中所有integer最大的值.
Space: O(n).
AC Java:
class Solution {
public int shoppingOffers(List<Integer> price, List<List<Integer>> special, List<Integer> needs) {
return dfs(price, special, needs);
} private int dfs(List<Integer> price, List<List<Integer>> special, List<Integer> needs){
// Direct purchase price
int res = directPurchase(price, needs);
for(List<Integer> offer : special){
if(!isValid(offer, needs)){
continue;
} List<Integer> restNeeds = new ArrayList<Integer>();
for(int i = 0; i<needs.size(); i++){
restNeeds.add(needs.get(i)-offer.get(i));
} res = Math.min(res, offer.get(offer.size()-1) + dfs(price, special, restNeeds));
} return res;
} private int directPurchase(List<Integer> price, List<Integer> needs){
int res = 0;
for(int i = 0; i<price.size(); i++){
res += price.get(i)*needs.get(i);
} return res;
} private boolean isValid(List<Integer> offer, List<Integer> needs){
for(int i = 0; i<needs.size(); i++){
if(offer.get(i) > needs.get(i)){
return false;
}
} return true;
}
}
LeetCode Shopping Offers的更多相关文章
- [LeetCode] Shopping Offers 购物优惠
In LeetCode Store, there are some kinds of items to sell. Each item has a price. However, there are ...
- LeetCode 638 Shopping Offers
题目链接: LeetCode 638 Shopping Offers 题解 dynamic programing 需要用到进制转换来表示状态,或者可以直接用一个vector来保存状态. 代码 1.未优 ...
- Leetcode之深度优先搜索&回溯专题-638. 大礼包(Shopping Offers)
Leetcode之深度优先搜索&回溯专题-638. 大礼包(Shopping Offers) 深度优先搜索的解题详细介绍,点击 在LeetCode商店中, 有许多在售的物品. 然而,也有一些大 ...
- LC 638. Shopping Offers
In LeetCode Store, there are some kinds of items to sell. Each item has a price. However, there are ...
- Week 9 - 638.Shopping Offers - Medium
638.Shopping Offers - Medium In LeetCode Store, there are some kinds of items to sell. Each item has ...
- 洛谷P2732 商店购物 Shopping Offers
P2732 商店购物 Shopping Offers 23通过 41提交 题目提供者该用户不存在 标签USACO 难度提高+/省选- 提交 讨论 题解 最新讨论 暂时没有讨论 题目背景 在商店中, ...
- poj 1170 Shopping Offers
Shopping Offers Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 4696 Accepted: 1967 D ...
- USACO 3.3 Shopping Offers
Shopping OffersIOI'95 In a certain shop, each kind of product has an integer price. For example, the ...
- HDU 1170 Shopping Offers 离散+状态压缩+完全背包
题目链接: http://poj.org/problem?id=1170 Shopping Offers Time Limit: 1000MSMemory Limit: 10000K 问题描述 In ...
随机推荐
- input propertyChange
結合 HTML5 標準事件 oninput 和 IE 專屬事件 onpropertychange 事件來監聽輸入框值變化. oninput 是 HTML5 的標準事件,對於檢測 textarea, i ...
- $GitHub边用边总结
以下用法是在ubuntu系统下的用法,主要内容整理自'廖雪峰的官方网站'. #1.安装git$ sudo apt-get install git #2.声明git账号$ git config --gl ...
- webAPI支持跨域
问题描述: 添加引用:右键项目→添加nuget包 在:App_Start/WebApiConfig.Register中添加如下一句话 //跨域配置 config.EnableCors(new Enab ...
- Android BlueDroid(蓝牙协议栈)
Android BlueDroid(一):BlueDroid概述 Android BlueDroid(二):BlueDroid蓝牙开启过程init Android BlueDroid(三):BlueD ...
- 【Head First Servlets and JSP】笔记5:HttpServletResponse resp
[HttpServletResponse resp] [由servlet处理响应] 1.一般可以用通过resp获得一个输出流(writer),然后通过输出流将HTML写入响应.例如: resp.set ...
- iOS_SDWebImage框架分析
SDWebImage 支持异步的图片下载+缓存,提供了 UIImageView+WebCacha 的 category,方便使用.使用SDWebImage首先了解它加载图片的流程. 入口 setIma ...
- SQL Server获取数据库的当前连接状态
SELECT * FROM [Master].[dbo].[SYSPROCESSES] WHERE [DBID]=(SELECT [DBID] FROM [Master].[dbo].[SYSDATA ...
- maven创建web工程Spring配置文件找不到
使用maven创建web工程,将Spring配置文件applicationContext.xml放在src/resource下,用eclipse编译时提示class path resource [ap ...
- 基于netty的异步http请求
package com.pt.utils; import io.netty.bootstrap.Bootstrap; import io.netty.channel.ChannelFuture; im ...
- 通过yum安装mysql
在linux中安装数据库首选MySQL,Mysql数据库的第一个版本就是发行在Linux系统上,其他选择还可以有postgreSQL,oracle等 在Linux上安装mysql数据库,我们可以去其官 ...