买卖股票Ⅲ

https://leetcode.cn/problems/best-time-to-buy-and-sell-stock-iii/

无语了。。。
写的很好就是怎么都过不了。。。

还是就用代码随想录的写法吧。。。

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        vector<vector<int>> dp(n, vector<int> (5, 0));
        dp[0][1] = -prices[0];
        dp[0][3] = -prices[0];
        for (int i = 1; i < n; i++) {
            dp[i][0] = dp[i - 1][0];
            dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] - prices[i]);
            dp[i][2] = max(dp[i - 1][2], dp[i - 1][1] + prices[i]);
            dp[i][3] = max(dp[i - 1][3], dp[i - 1][2] - prices[i]);
            dp[i][4] = max(dp[i - 1][4], dp[i - 1][3] + prices[i]);
        }

        return *max_element(dp.back().begin(), dp.back().end());  
    }
};

买卖股票Ⅳ

https://leetcode.cn/problems/best-time-to-buy-and-sell-stock-iv/description/

根据Ⅲ找规律：

class Solution {
public:
    int maxProfit(int k, vector<int>& prices) {
        int n = prices.size();
        vector<vector<int>> dp(n, vector<int>(2 * k + 1, 0));
        for (int i = 0; i < 2*k+1; i++) {
            if (i % 2 != 0) dp[0][i] = -prices[0];
        }
        for (int i = 1; i < n; i++) {
            dp[i][0] = dp[i - 1][0];
            for (int j = 1; j < 2*k+1; j++) {
                if (j % 2 != 0) dp[i][j] = max(dp[i-1][j], dp[i-1][j-1] - prices[i]);
                else dp[i][j] = max(dp[i-1][j], dp[i-1][j-1] + prices[i]);
            }
        }

        return *max_element(dp.back().begin(), dp.back().end());  
    }
};