LeetCode-122-买卖股票的最佳时机 II

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解题思路1 (动态规划)：

1. 第i天没持有股票最大收益由两个状态转移来: 保持i-1天的状态，继续不持有，或者i-1天持有，第i天卖出； dp_0[i] = max(dp_0[i], dp_1[i-1]+prices[i])
2. 第i天持有股票最大收益有两个状态转移来：保持i-1天的状态，持有股票，或者i-1没持有，第i天买入； dp_1[i] = max(dp_1[i-1], dp_0[i]-prices[i])

python3代码如下：

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        len1 = len(prices)
        dp_0 = [0 for _ in range(len1)]
        dp_1 = [0 for _ in range(len1)]
        dp_1[0] = -prices[0]

        for i in range(1, len(prices)):
            dp_0[i] = max(dp_0[i-1], dp_1[i-1]+prices[i])
            dp_1[i] = max(dp_1[i-1], dp_0[i-1]-prices[i])

        return dp_0[-1]

解题思路2：

1. 收益累加，假如比前一天数值大，则累加入总收益

python3代码如下：

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        maxProfit = 0 
        for i in range(1, len(prices)):
            if prices[i] > prices[i-1]:
                maxProfit += prices[i] - prices[i-1]
        return maxProfit

解题思路3：

1. 在思路1基础上的优化，取代了数组存储状态，只保留当天收益

python3代码如下：

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        len1 = len(prices)
        dp_0 = 0
        dp_1 = -prices[0]

        for i in range(1, len(prices)):
            dp_0 = max(dp_0, dp_1+prices[i])
            dp_1 = max(dp_1, dp_0-prices[i])
        return dp_0