123. Best Time to Buy and Sell Stock III

You are given an array prices where prices[i] is the price of a given stock on the ith day.

Find the maximum profit you can achieve. You may complete at most two transactions.

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

题目解析：

• 最多两次交易，所以可以分为五种状态：初始状态，第一次持有，第一次卖出，第二次持有，第二次卖出
• 定义二维数组：第一维是天数，第二维是状态
• 当前天的状态只与前一天的状态有关，而且当前状态只与前一个状态有关，所以可以使用一维数组

代码：

// 二维数组
class Solution {
    public int maxProfit(int[] prices) {
        if (prices.length == 1) return 0;
        // 五种状态：初始状态，第一次持有，第一次卖出，第二次持有，第二次卖出
        int[][] dp = new int[prices.length][5];
        dp[0][1] = -prices[0];
        dp[0][3] = -prices[0];
        for (int i = 1; i < prices.length; i++) {
            dp[i][1] = Math.max(dp[i-1][1], dp[i-1][0] - prices[i]);
            dp[i][2] = Math.max(dp[i-1][2], dp[i-1][1] + prices[i]);
            dp[i][3] = Math.max(dp[i-1][3], dp[i-1][2] - prices[i]);
            dp[i][4] = Math.max(dp[i-1][4], dp[i-1][3] + prices[i]);
        }
        return dp[prices.length-1][4];
    }
}

// 一维数组
class Solution {
    public int maxProfit(int[] prices) {
        if (prices.length == 1) return 0;
        // 五种状态：初始状态，第一次持有，第一次卖出，第二次持有，第二次卖出
        int[] dp = new int[5];
        dp[1] = -prices[0];
        dp[3] = -prices[0];
        for (int i = 1; i < prices.length; i++) {
            dp[1] = Math.max(dp[1], dp[0] - prices[i]);
            dp[2] = Math.max(dp[2], dp[1] + prices[i]);
            dp[3] = Math.max(dp[3], dp[2] - prices[i]);
            dp[4] = Math.max(dp[4], dp[3] + prices[i]);
        }
        return dp[4];
    }
}

188. Best Time to Buy and Sell Stock IV

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

Find the maximum profit you can achieve. You may complete at most k transactions: i.e. you may buy at most k times and sell at most k times.

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

题目解析：

• 与上一题类似

代码：

class Solution {
    public int maxProfit(int k, int[] prices) {
        if (prices.length == 1) return 0;
        int[] dp = new int[k * 2 + 1];
        for (int i = 0; i < k * 2 + 1; i++) {
            if (i % 2 == 1) dp[i] = -prices[0];
        }
        for (int i = 1; i < prices.length; i++) {
            for (int j = 1; j < k * 2 + 1; j++) {
                if (j % 2 == 0) {
                    dp[j] = Math.max(dp[j], dp[j - 1] + prices[i]);
                } else {
                    dp[j] = Math.max(dp[j], dp[j - 1] - prices[i]);
                }
                
            }
        }
        return dp[k * 2];
    }
}