代码随想录算法训练营Day51 | 309.最佳买卖股票时机含冷冻期，714.买卖股票的最佳时机含手续费，总结

309.最佳买卖股票时机含冷冻期

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C++解法

class Solution {
public:
    int maxProfit(vector& prices) {
        int n = prices.size();
        if (n == 0) return 0;
        vector> dp(n, vector(4, 0));
        dp[0][0] -= prices[0];
        for (int i = 1; i < n; i++) {
            dp[i][0] = max(dp[i - 1][0], max(dp[i - 1][3] - prices[i], dp[i - 1][1] - 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[n - 1][3], max(dp[n - 1][1], dp[n - 1][2]));
    }
};

Python解法

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        n = len(prices)
        if n == 0:
            return 0
        dp = [[0]*4 for _ in range(len(prices))]
        dp[0][0] = -prices[0]
        for i in range(1, n):
            dp[i][0] = max(dp[i-1][0], max(dp[i-1][3], dp[i-1][1] - 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-2][2]
        return max(dp[n-1][3], dp[n-1][1], dp[n-1][2])

714.买卖股票的最佳时机含手续费

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C++解法

class Solution {
public:
    int maxProfit(vector& prices, int fee) {
        int n = prices.size();
        vector> dp(n, vector(2, 0));
        dp[0][0] -= prices[0]; 
        for (int i = 1; i < n; 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 max(dp[n - 1][0], dp[n - 1][1]);
    }
};

Python解法

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

总结

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