714 Best Time to Buy and Sell Stock with Transaction Fee

You are given an array prices where prices[i] is the price of a given stock on the ith day, and an integer fee representing a transaction fee.

Find the maximum profit you can achieve. You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction.

Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

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.

Example 2:

Input: prices = [1,3,7,5,10,3], fee = 3
Output: 6

Constraints:

• 1 < prices.length <= 5 * 104

• 0 < prices[i], fee < 5 * 104

自己想了半天，没想出来。然后看了hint我也觉得是DP。还是没想出来。看了答案，觉得自己是想不出来的。这里的递推公式是，如果你今天有股票，你要不就是昨天hold到现在，要不就是今天刚买。所以max(hold, cash - prices[i])，买的时候不用给fee。如果你今天没股票，你要不就是昨天开始就没了，或者你昨天有今天刚卖掉。所以是max(cash, hold + prices[i] - fee)。T:O(N) S:O(1)

public int maxProfit(int[] prices, int fee) {
    if (prices == null || prices.length == 0) {
        return 0;
    }

    int hold = -prices[0];
    int cash = 0;
    for (int i = 1; i < prices.length; i++) {
        hold = Math.max(hold, cash - prices[i]);
        cash = Math.max(cash, hold + prices[i] - fee);
    }

    return cash;
}