[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;
    }
};

　　

　　

 

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