第一题、买卖股票的最佳时机含冷冻期 力扣题目链接

class Solution {
public:int maxProfit(vector<int>& prices) {if(prices.size() == 0) return 0;vector<vector<int>> dp(prices.size(), vector<int>(4, 0));dp[0][0] = -prices[0];dp[0][1] = dp[0][2] = dp[0][3] = 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]));  // 持有状态dp[i][1] = max(dp[i-1][1], dp[i-1][3]);  // 保持卖出状态dp[i][2] = dp[i-1][0] + prices[i];  // 具体卖出操作dp[i][3] = dp[i-1][2];  // 冷冻期}return max(dp[prices.size() - 1][1], max(dp[prices.size() - 1][2], dp[prices.size() - 1][3]));}
};

第二题、买卖股票的最佳时机含手续费 力扣题目链接

class Solution {
public:int maxProfit(vector<int>& prices, int fee) {if(prices.size() == 0) return 0 ;vector<vector<int>> dp(prices.size(), vector<int>(2, 0));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];}
};

买卖股票问题总结：