leetcode 714. Best Time to Buy and Sell Stock with Transaction Fee | 714. 买卖股票的佳最时机含手续费（递归-＞傻缓存-＞dp）

题目

https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/

题解

经典的 暴力递归 -> 傻缓存 -> DP，重要思路来源于之前做过的 leetcode 309. Best Time to Buy and Sell Stock with Cooldown | 309. 最佳买卖股票时机含冷冻期（动态规划）

当年看答案也摸不清门路的 dp，现在应不是玄学了，哈哈。

将「买入」和「卖出」分开进行考虑：「买入」为负收益，而「卖出」为正收益。在初入股市时，你只有「买入」的权利，只能获得负收益。而当你「买入」之后，你就有了「卖出」的权利，可以获得正收益。显然，我们需要尽可能地降低负收益而提高正收益，因此我们的目标总是将收益值最大化。因此，我们可以使用动态规划的方法，维护在股市中每一天结束后可以获得的「累计最大收益」，并以此进行状态转移，得到最终的答案。

没有草稿，直接上代码：

class Solution {public int maxProfit(int[] prices, int fee) {// Approach 1: Recursive, Brute Force
//        return process1(prices, 0, false,fee);// Approach 2: Recursion with Memoization
//        int[][] dp = new int[2][prices.length + 1];
//        Arrays.fill(dp[0], Integer.MIN_VALUE);
//        Arrays.fill(dp[1], Integer.MIN_VALUE);
//        dp[0][prices.length] = 0;
//        dp[1][prices.length] = 0;
//        return process2(prices, 0, 0, fee, dp);// Approach 3: Dynamic Programmingint[][] dp = new int[2][prices.length + 1];Arrays.fill(dp[0], Integer.MIN_VALUE);Arrays.fill(dp[1], Integer.MIN_VALUE);dp[0][prices.length] = 0;dp[1][prices.length] = 0;for (int i = prices.length - 1; i >= 0; i--) {for (int pending = 0; pending <= 1; pending++) {int p1 = Integer.MIN_VALUE;int p2 = Integer.MIN_VALUE;int p3 = Integer.MIN_VALUE;int p4 = Integer.MIN_VALUE;if (pending == 1) {p1 = dp[1][i + 1]; // 不卖p2 = dp[0][i + 1] + prices[i] - fee; // 卖} else {p3 = dp[1][i + 1] - prices[i]; // 买p4 = dp[0][i + 1]; // 不买}dp[pending][i] = Math.max(Math.max(p1, p2), Math.max(p3, p4));}}return dp[0][0];}// 从i位置开始做决策，获得的最大收益（pending表示当前是否持有股票）
//    public int process1(int[] prices, int i, boolean pending, int fee) {//        if (i == prices.length) return 0;
//
//        int p1 = Integer.MIN_VALUE;
//        int p2 = Integer.MIN_VALUE;
//        int p3 = Integer.MIN_VALUE;
//        int p4 = Integer.MIN_VALUE;
//        if (pending) {// 持有股票
//            p1 = process1(prices, i + 1, true, fee); // 不卖
//            p2 = process1(prices, i + 1, false, fee) + prices[i] - fee; // 卖
//        } else { // 未持有股票
//            p3 = process1(prices, i + 1, true, fee) - prices[i]; // 买
//            p4 = process1(prices, i + 1, false, fee); // 不买
//        }
//        return Math.max(Math.max(p1, p2), Math.max(p3, p4));
//    }//    public int process2(int[] prices, int i, int pending, int fee, int[][] dp) {//        if (dp[pending][i] != Integer.MIN_VALUE) return dp[pending][i];
//
//        int p1 = Integer.MIN_VALUE;
//        int p2 = Integer.MIN_VALUE;
//        int p3 = Integer.MIN_VALUE;
//        int p4 = Integer.MIN_VALUE;
//        if (pending == 1) {// 持有股票
//            p1 = process2(prices, i + 1, 1, fee, dp); // 不卖
//            p2 = process2(prices, i + 1, 0, fee, dp) + prices[i] - fee; // 卖
//        } else { // 未持有股票
//            p3 = process2(prices, i + 1, 1, fee, dp) - prices[i]; // 买
//            p4 = process2(prices, i + 1, 0, fee, dp); // 不买
//        }
//
//        int result = Math.max(Math.max(p1, p2), Math.max(p3, p4));
//        dp[pending][i] = result;
//        return result;
//    }
}

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leetcode 714. Best Time to Buy and Sell Stock with Transaction Fee | 714. 买卖股票的佳最时机含手续费（递归-＞傻缓存-＞dp）

题目

题解

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