代码随想录训练营day42|动态规划part9
买卖股票的最佳时机IV
力扣题目链接
class Solution {
public:int maxProfit(int k, vector<int>& prices) {vector<vector<int>> dp(prices.size(), vector<int>(2*k+1, 0));for(int i = 0; i < k; i++){dp[0][2*i+1] = -prices[0];}for(int i = 1; i < prices.size(); i++){for(int j = 0; j < k; j++){dp[i][2*j+1] = max(dp[i-1][2*j+1], dp[i][2*j]-prices[i]);dp[i][2*j+2] = max(dp[i-1][2*j+2], dp[i][2*j+1]+prices[i]);}}return dp[prices.size()-1][2*k];}
};
最佳买卖股票时机含冷冻期
力扣题目链接
class Solution {
public:int maxProfit(vector<int>& prices) {// 买（0）--卖（1）--无操作（2）int n = prices.size();vector<vector<int>> dp(n, vector<int>(3,0));dp[0][0] = -prices[0];for(int i = 1; i < n; i++){dp[i][0] = max(dp[i-1][0], dp[i-1][2] - prices[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]);}return max(dp[n-1][1], dp[n-1][2]);}
};
买卖股票的最佳时机含手续费
力扣题目链接
class Solution {
public:int maxprofit_all(vector<int>& prices, int fee) {int n = prices.size();//尝试了不用dp数组，只用两个变量int profit_parts = -prices[0]; //持有股票的收入int profit_all = 0; //没有股票的收入for(int i = 1; i < n; i++){profit_parts = max(profit_parts, profit_all - prices[i]);profit_all = max(profit_all, profit_parts + prices[i] - fee);}return profit_all;}
};