代码随想录算法训练营 day50| 123.买卖股票的最佳时机III、188.买卖股票的最佳时机IV

一、123.买卖股票的最佳时机III

力扣题目链接

增加了两次的限制，相应的就是需要考虑的状态改变

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

二、188.买卖股票的最佳时机IV

力扣题目链接

class Solution {
public:
    int maxProfit(int k, vector& prices) {

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