LWC 55：714. Best Time to Buy and Sell Stock with Transaction Fee
传送门：714. Best Time to Buy and Sell Stock with Transaction Fee
Problem:
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.
思路： 动态规划，对于每一个可能的买入点都有三种状态：买，卖，什么都不做，枚举出这些状态的组合，求最大profit即可。
目标：
- 前i天的持有股票数最大，记为：hold[i]
- 前i天的股票利益最大，记为：unHold[i]
状态转移：
对于第i 1天，三种状态，买，卖，什么都不做。
买： hold[i] = max{unHold[i] - prices[i 1] - fee, hold[i - 1];
卖：unHold[i] = max{hold[i] prices[i 1], unHold[i - 1];
代码如下：
代码语言：javascript复制 public int maxProfit(int[] prices, int fee) {
int n = prices.length;
int[] hold = new int[n 1];
int[] unHold = new int[n 1];
hold[0] = Integer.MIN_VALUE;
for (int i = 1; i <= n; i) {
hold[i] = Math.max(hold[i - 1], unHold[i - 1] - prices[i - 1] - fee);
unHold[i] = Math.max(unHold[i - 1], hold[i - 1] prices[i - 1]);
}
return unHold[n];
}
