[LeetCode] 714. Best Time to Buy and Sell Stock with Transaction Fee 买卖股票的最佳时间有交易费
Your are given an array of integers prices
, for which the i
-th element is the price of a given stock on day i
; and a non-negative integer fee
representing a transaction fee.
You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction. You may not buy more than 1 share of a stock at a time (ie. you must sell the stock share before you buy again.)
Return the maximum profit you can make.
Example 1:
Input: prices = [1, 3, 2, 8, 4, 9], fee = 2
Output: 8
Explanation: The maximum profit can be achieved by:
- Buying at prices[0] = 1
- Selling at prices[3] = 8
- Buying at prices[4] = 4
- Selling at prices[5] = 9
The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.
Note:
0 < prices.length <= 50000
.0 < prices[i] < 50000
.0 <= fee < 50000
.
还是买卖股票的最佳时间问题,这题每一次买卖时会有交易费。
解法:DP。第i天的利润分成两个,用两个dp数组分别进行计算,buy[i], sell[i]。
初始值:buy[0]=-prices[0], sell[0]=0
公式:
buy[i] = Math.max(buy[i - 1], sell[i - 1] - prices[i])
第i天买,如果第i-1天是买,就不能买了,利润是buy[i-1]。如果i-1天是卖,就可以买,利润是sell[i-1] - prices[i]。
sell[i] = Math.max(sell[i - 1], buy[i - 1] + prices[i])
第i天卖,如果第i-1天是卖,就不能卖了,利润是sell[i-1]。如果i-1天是买,就可以卖,利润是buy[i - 1] + prices[i]。
Most consistent ways of dealing with the series of stock problems
Java: pay the fee when buying the stock
public int maxProfit(int[] prices, int fee) {
if (prices.length <= 1) return 0;
int days = prices.length, buy[] = new int[days], sell[] = new int[days];
buy[0]=-prices[0]-fee;
for (int i = 1; i<days; i++) {
buy[i] = Math.max(buy[i - 1], sell[i - 1] - prices[i] - fee); // keep the same as day i-1, or buy from sell status at day i-1
sell[i] = Math.max(sell[i - 1], buy[i - 1] + prices[i]); // keep the same as day i-1, or sell from buy status at day i-1
}
return sell[days - 1];
}
Java: pay the fee when selling the stock
public int maxProfit(int[] prices, int fee) {
if (prices.length <= 1) return 0;
int days = prices.length, buy[] = new int[days], sell[] = new int[days];
buy[0]=-prices[0];
for (int i = 1; i<days; i++) {
buy[i] = Math.max(buy[i - 1], sell[i - 1] - prices[i]); // keep the same as day i-1, or buy from sell status at day i-1
sell[i] = Math.max(sell[i - 1], buy[i - 1] + prices[i] - fee); // keep the same as day i-1, or sell from buy status at day i-1
}
return sell[days - 1];
}
Python:
class Solution(object):
def maxProfit(self, prices, fee):
"""
:type prices: List[int]
:type fee: int
:rtype: int
"""
cash, hold = 0, -prices[0]
for i in xrange(1, len(prices)):
cash = max(cash, hold+prices[i]-fee)
hold = max(hold, cash-prices[i])
return cash
C++:
class Solution {
public:
int maxProfit(vector<int>& prices, int fee) {
int s0 = 0, s1 = INT_MIN;
for(int p:prices) {
int tmp = s0;
s0 = max(s0, s1+p);
s1 = max(s1, tmp-p-fee);
}
return s0;
}
};
All LeetCode Questions List 题目汇总
[LeetCode] 714. Best Time to Buy and Sell Stock with Transaction Fee 买卖股票的最佳时间有交易费的更多相关文章
- Week 7 - 714. Best Time to Buy and Sell Stock with Transaction Fee & 718. Maximum Length of Repeated Subarray
714. Best Time to Buy and Sell Stock with Transaction Fee - Medium Your are given an array of intege ...
- 714. Best Time to Buy and Sell Stock with Transaction Fee
问题 给定一个数组,第i个元素表示第i天股票的价格,可执行多次"买一次卖一次",每次执行完(卖出后)需要小费,求最大利润 Input: prices = [1, 3, 2, 8, ...
- 【LeetCode】714. Best Time to Buy and Sell Stock with Transaction Fee 解题报告(Python & C++)
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 动态规划 日期 题目地址:https://leetc ...
- 【leetcode】714. Best Time to Buy and Sell Stock with Transaction Fee
题目如下: Your are given an array of integers prices, for which the i-th element is the price of a given ...
- 714. Best Time to Buy and Sell Stock with Transaction Fee有交易费的买卖股票
[抄题]: Your are given an array of integers prices, for which the i-th element is the price of a given ...
- Leetcode之动态规划(DP)专题-714. 买卖股票的最佳时机含手续费(Best Time to Buy and Sell Stock with Transaction Fee)
Leetcode之动态规划(DP)专题-714. 买卖股票的最佳时机含手续费(Best Time to Buy and Sell Stock with Transaction Fee) 股票问题: 1 ...
- LeetCode-714.Best Time to Buy and Sell Stock with Transaction Fee
Your are given an array of integers prices, for which the i-th element is the price of a given stock ...
- [LeetCode] Best Time to Buy and Sell Stock with Transaction Fee 买股票的最佳时间含交易费
Your are given an array of integers prices, for which the i-th element is the price of a given stock ...
- LeetCode Best Time to Buy and Sell Stock with Transaction Fee
原题链接在这里:https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/descripti ...
随机推荐
- @Scope("prototype")
spring中bean的scope属性,有如下5种类型: singleton 表示在spring容器中的单例,通过spring容器获得该bean时总是返回唯一的实例prototype表示每次获得bea ...
- websocket简单样例
服务端 var server = require('ws').Server; }); serv.on('connection',function(socket){ socket.send('hello ...
- ActiveMQ基础
消息队列的作用 为什么使用ActiveMQ,不使用其他工具 下载安装包并启动 http://localhost:8161/admin/ (账号:admin:admin) Java实现步骤: // 1. ...
- 微信小程序——选择某个区间的数字
很久没有更新文章啦~~记录下今天弄的一个小功能. 先上图: 需求很简单: 第1列改变的时候,第2列也随着改变,并且比第1列大1k. 这里用到了微信的picker 组件,对于不太熟练这个组件的小伙伴可以 ...
- c++实用语法
数组的快捷初始化 int inq[110] memset(inq, 0, sizeof(inq)); string到char数组的转换: string str ("Please split ...
- python -- 连接 orclae cx_Oracle的使用 二
转:https://www.cnblogs.com/cyxiaer/p/9396861.html 必需的Oracle链接库的下载地址:https://www.oracle.com/technetwor ...
- C语言博客作业00--我的第一篇博客
1.你对网络专业或者计算机专业了解是怎样? 起初 起初对于我来说,计算机专业毕业后就相当于程序员,或者去开发一些游戏,软件等等,而学得特别优秀的可能会成为黑客,就像电影电视剧里演得那样,这是我一开始的 ...
- 在Matlab中的tick可以调整方向
需要将axis对话框的More property打开,修改TickDir,可从In改成Out.
- 获取页面scroll高度
记录一下获取 scroll 高度的方法 经实际测试: document.body.scrollTop 在 chrome 下会返回0. 所以 document.documentElement.scrol ...
- 小数据池/is和==/再谈编码作业
# 1,老男孩好声选秀大赛评委在打分的时候呢, 可以输入分数. 假设, 老男孩有10个评委. 让10个评委进行打分, 要求, 分数必须高于5分, 低于10分.将每个评委的打分情况保存在列表中. pin ...