代码随想录算法训练营Day50 | 123.买卖股票的最佳时机III，188.买卖股票的最佳时机IV

123.买卖股票的最佳时机III

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C++解法

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if (prices.size() == 0) return 0;
        vectorint>> dp(prices.size(), vector<int>(5, 0));
        dp[0][1] = -prices[0];
        dp[0][3] = -prices[0];
 
        for (int i = 1; i < prices.size(); i++){
            dp[i][0] = dp[i - 1][0];
            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[prices.size() - 1][4];
    }
};

Python解法

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        if len(prices) == 0:
            return 0
        dp = [[0] * 5 for _ in range(len(prices))]
        dp[0][1] = -prices[0]
        dp[0][3] = -prices[0]
        for i in range(1, len(prices)):
            dp[i][0] = dp[i - 1][0]
            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[-1][4]

188.买卖股票的最佳时机IV 

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C++解法

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

Python解法

class Solution:
    def maxProfit(self, k: int, prices: List[int]) -> int:
        if len(prices) == 0:
            return 0
        dp = [[0] * (2*k+1) for _ in range(len(prices))]
        for j in range(1, 2*k, 2):
            dp[0][j] = -prices[0]
        for i in range(1, len(prices)):
            for j in range(0, 2*k-1, 2):
                dp[i][j+1] = max(dp[i-1][j+1], dp[i-1][j] - prices[i])
                dp[i][j+2] = max(dp[i-1][j+2], dp[i-1][j+1] + prices[i])
        return dp[-1][2*k]