leetcode -- 714. Best Time to Buy and Sell Stock with Transaction Fee

题目描述

题目难度：Medium

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.

AC代码

也是利用贪心算法求解。
参考：https://blog.youkuaiyun.com/liukcqu/article/details/82463294
对于第i天来说，手中要么没有股票，要么手持一个股票。用sold表示第i天没有股票的收益，hold表示第i天手持股票的收益。
没有股票的情况：要么是前一天就没有股票，要么是今天卖掉了，所以这种情况下，最大值就是比较这两者。sold[i] = max(sold[i-1],hold[i-1]+price[i]-fee)
手持股票的情况：要么是前一天就持有股票，要么今天买的股票，取其中最大值。
hold[i] = max(hold[i-1],sold[i-1] - price[i])

class Solution {
   public int maxProfit(int[] prices, int fee) {       
        int preHold= -prices[0];
        int preSold = 0;
        
        for(int i = 1; i < prices.length; i++) {
            int hold = Math.max(preSold- prices[i], preHold);
            int sold = Math.max(preHold+ prices[i] - fee, preSold);
            preHold = hold ;
            preSold = sold ;
        }
        return preSold;
    }
}