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

Description

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.

解题思路

采用动态规划，状态方程为（第i天）
buy[i] = max(buy[i - 1], sell[i - 1] - prices[i]);
sell[i] = max(sell[i - 1], buy[i - 1] + prices[i]);
返回结果为sell[last_day]

Solution I – pay the fee when buying the stock:

class Solution {public:    int maxProfit(vector<int>& prices, int fee) {        if(prices.size() <= 1) return 0;        int size = prices.size();        vector<int>buy(size+1, INT_MIN);        vector<int>sell(size+1, 0);        for (int i = 0; i < size;i++) {            buy[i+1] = max(buy[i],sell[i]-prices[i]-fee);            sell[i+1] = max(sell[i],buy[i+1]+prices[i]);        }        return sell.back();    }};

Solution II – pay the fee when selling the stock:

class Solution {public:    int maxProfit(vector<int>& prices, int fee) {        if(prices.size() <= 1) return 0;        int size = prices.size();        vector<int>buy(size+1, INT_MIN);        vector<int>sell(size+1, 0);        for (int i = 0; i < size;i++) {            buy[i+1] = max(buy[i],sell[i]-prices[i]);            sell[i+1] = max(sell[i],buy[i+1]-fee+prices[i]);        }        return sell.back();    }};