刷题笔记50 动态规划 part11

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目录

• 动态规划
  • 123.买卖股票的最佳时机III
  • 188.买卖股票的最佳时机IV

动态规划

● 123.买卖股票的最佳时机III
● 188.买卖股票的最佳时机IV

123.买卖股票的最佳时机III

123.买卖股票的最佳时机III

法1：动态规划

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

法2：滚动数组

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

188.买卖股票的最佳时机IV

188.买卖股票的最佳时机IV
法1：动态规划

       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 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];
    }