Day 50 | 123. Best Time to Buy and Sell Stock III | 188. Best Time to Buy and Sell Stock IV

股票买卖策略动态规划解题概览,

原创已于 2023-08-16 23:20:06 修改 · 137 阅读

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于 2023-08-01 16:18:18 首次发布

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文章介绍了一系列编程问题，涵盖二分搜索、数组操作、链表、树遍历、股票买卖策略等，重点讲解了动态规划在BestTimetoBuyandSellStock问题中的应用及其扩展版本。

123. Best Time to Buy and Sell Stock III

Question Link

class Solution {
    public int maxProfit(int[] prices) {
        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];
    }
}

Meaning of dp array
- dp[i][0]: no operation(In fact, we can also not set this state)
- dp[i][1]: maximum cash of first holding the stock on day i
- dp[i][2]: maximum cash of first doesn’t holding stocks on day i
- dp[i][3]: maximum cash of second holding the stock on the day i
- dp[i][4]: maximum cash of second holding the stock on the day i
- The hold here doesn’t mean buy today. It is also possible that I bought it yesterday and keep it today.
Recursion Formula:
- If hold the stock on i-1 day, let’s keep it that way.
- If first buy stock on day i, current cash is dp[i - 1][0] - prices[i]
- So the dp[i][1] should choose the larger one: dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] - prices[i])
- If doesn’t hold the the stock on i - 1, let’s keep it that way.
- If first sell stock on day i, current cash is prices[i] + dp[i - 1][1]
- Again dp[i][2] takes the larger one: dp[i][2] = max(dp[i - 1][2], dp[i - 1][1] + prices[i])
- So dp[i][3] and dp[i][4] are the similar like above. The complete recursion formulas are as follows.
  - 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])
Initialize the DP array
- According to the recursion formula, the values of dp array are derived from dp[0][0], dp[0][1], dp[0][2], dp[0][3] and dp[0][4]
- dp[0][0] = 0: no operation
- dp[0][1] = -prices[0]: buy the stock on day 0
- dp[0][2] = 0: first sell the stock on day 0
- dp[0][3] = -prices[0]: second buy the stock on the day 0
- dp[0][4]= 0: second sell the stock on the day 0
Traversal Order is from front to back.

188. Best Time to Buy and Sell Stock IV

Question Link

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

Compare this question to the 123. Best Time to Buy and Sell Stock III, the most significant difference is buying the stock when k is odd and selling the stock when k is even.