代码随想录刷题60Day

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目录

前言

买卖股票的最佳时机1

买卖股票的最佳时机2

买卖股票的最佳时机3

买卖股票的最佳时机4

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前言

本日着重于多状态问题的处理，各状态之间会有一定联系，状态转移方程将不再局限一个。

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买卖股票的最佳时机1

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

买卖股票的最佳时机2

    int maxProfit1(vector<int>& prices) {const int size = prices.size();vector<vector<int>> dp(2, vector<int> (2, 0));dp[0][1] = -prices[0];for (int i = 1; i < size; ++i){dp[i % 2][0] = max(dp[(i - 1) % 2][0], dp[(i - 1) % 2][1] + prices[i]);dp[i % 2][1] = max(dp[(i - 1) % 2][1], dp[(i - 1) % 2][0] - prices[i]);}return dp[(size - 1) % 2][0];}

买卖股票的最佳时机3

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

买卖股票的最佳时机4

int maxProfit(int k, vector<int>& prices){//----------------case 1------------------------const int size = prices.size();vector<vector<int>> dp(size, vector<int>(2 * k + 1, 0));for (int i = 1; i <= k; ++i)dp[0][i * 2 - 1] = -prices[0];for (int i = 1; i < size; ++i){int sign = -1;for (int j = 1; j < (2 * k + 1); ++j){dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - 1] + sign * prices[i]);sign *= -1;}}return dp[size - 1][2 * k];//----------------case 2-----------------------const int size = prices.size();vector<vector<int>> dp(2, vector<int>(2 * k + 1, 0));for (int i = 1; i <= k; ++i)dp[0][i * 2 - 1] = -prices[0];for (int i = 1; i < size; ++i){int sign = -1;for (int j = 1; j < (2 * k + 1); ++j){dp[i % 2][j] = max(dp[(i - 1) % 2][j], dp[(i - 1) % 2][j - 1] + sign * prices[i]);sign *= -1;}}return dp[(size - 1) % 2][2 * k];}