Leetcode Problem 417

2018/12/9

714. Best Time to Buy and Sell Stock with Transaction Fee

問題描述

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.

測試樣例

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.

問題分析

本題難度爲Medium！屬於動態規劃問題，已給出的函數定義爲

class Solution:
    def maxProfit(self, prices, fee):
        """
        :type prices: List[int]
        :type fee: int
        :rtype: int
        """

本題可以採用動態規劃的解決辦法；
dp[i][0]：第i天手中沒有股票時能賺的最高價；
dp[i][1]：第i天手中有股票時所賺的最高價；

當第i天手裏沒有股票時，這一天有兩個情況，第一種是在今天賣出了股票，從而一共賺取了dp[i-1][1]+prices[i]-fee，第二種情況是昨天就沒有股票，今天只是仍然沒有股票而已，這個時候dp[i][0]=dp[i-1][0];同理可以推導出dp[i][1]的轉移方程;
最終答案就是dp[n][0]，因爲必然要讓手裏的錢更多，肯定是手裏沒有股票的時候。
轉移方程爲：
dp[i][0] = max(dp[i-1][1]+prices[i]-fee, dp[i-1][0]);
dp[i][1] = max(dp[i-1][0]-prices[i], dp[i-1][1]);

代碼實現

python3

# coding: utf-8
class Solution:
    def maxProfit(self, prices, fee):
        """
        :type prices: List[int]
        :type fee: int
        :rtype: int
        """
        if len(prices) == 0: return 0
        dp = [[0,0] for i in range(len(prices))]
        dp[0][0] = 0
        dp[0][1] = -prices[0]
        for i in range(1, len(prices)):
            dp[i][0] = max(dp[i-1][0], dp[i-1][1]+prices[i]-fee)
            dp[i][1] = max(dp[i-1][1], dp[i-1][0]-prices[i])
        return dp[-1][0]