代码随想录算法打卡第四十四天, 新手自我记录一下刷题历程, 仅为自我打卡使用.

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309. 最佳买卖股票时机含冷冻期

class Solution {
public:int maxProfit(vector<int>& prices) {if (prices.size() < 2) return 0;vector<vector<int>> dp(prices.size(), {0, 0, 0, 0});dp[0][0] = -prices[0];dp[0][1] = 0;dp[0][2] = 0;dp[0][3] = 0;// yes // sold // today sale // frozenfor (int i = 1; i < prices.size(); ++i) {dp[i][0] = max(dp[i - 1][0], max(dp[i - 1][1] - prices[i], dp[i - 1][3] - 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[prices.size() - 1][1], max(dp[prices.size() - 1][2], dp[prices.size() - 1][3]));}
};

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714. 买卖股票的最佳时机含手续费

class Solution {
public:int maxProfit(vector<int>& prices, int fee) {// buy // have // no // vector<vector<int>> dp(prices.size(), vector<int>(3, 0));dp[0][0] = -prices[0] - fee;dp[0][1] = -prices[0] - fee;dp[0][2] = 0;for (int i = 1; i < prices.size(); ++i) {dp[i][0] = dp[i - 1][2] - prices[i] - fee;dp[i][1] = max(dp[i - 1][0], dp[i - 1][1]);dp[i][2] = max(dp[i - 1][2], max(dp[i - 1][0] + prices[i], dp[i - 1][1] + prices[i]));}return dp.back()[2];}
};

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股票系列要结束了, 撒花!

希望还能再坚持一天!