dp09

188. 买卖股票的最佳时机 IV

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

309. 买卖股票的最佳时机含冷冻期

2种状态

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if (prices.size() == 1) return 0;
        vector<vector<int>> dp(prices.size(), vector<int>(2, 0)); //1 持有 0 不持有
        dp[0][1] = -prices[0];
        dp[1][0] = max(dp[0][0], dp[0][1] + prices[1]);
        dp[1][1] = max(dp[0][0] - prices[1], dp[0][1]);
        for (int i = 2; i < prices.size(); i++) {
            dp[i][0] = max(dp[i - 1][0], dp[i - 1][1] + prices[i]);
            dp[i][1] = max(dp[i - 1][1], dp[i - 2][0] - prices[i]);
        }
        return max(dp[prices.size() - 1][0], dp[prices.size() - 1][1]);
    }
};

4种状态

        dp[0][0] = -prices[0];
        for (int i = 1; i < prices.size(); i++) {
            dp[i][0] = max(dp[i - 1][0], max(dp[i - 1][3] - prices[i], dp[i - 1][1] - prices[i]));
            dp[i][1] = max(dp[i - 1][1], dp[i - 1][3]);
            dp[i][2] = dp[i - 1][0] + prices[i]; 
            dp[i][3] = dp[i - 1][2];
        }
        return max(dp[n - 1][3], max(dp[n - 1][1], dp[n - 1][2]));

714. 买卖股票的最佳时机含手续费

class Solution {
public:
    int maxProfit(vector<int>& prices, int fee) {
        vector<vector<int>> dp(prices.size(), vector<int>(2, 0)); //0持有 1不持有
        dp[0][0] = -prices[0];
        for (int i = 1; i < prices.size(); i++) {
            dp[i][0] = max(dp[i - 1][0], dp[i - 1][1] - prices[i]);
            dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] + prices[i] - fee);
        }
        return dp[prices.size() - 1][1];
    }
};