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;
}
};