算法训练营day52

题目1：123. 买卖股票的最佳时机 III - 力扣（LeetCode）

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

题目2：188. 买卖股票的最佳时机 IV - 力扣（LeetCode）

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