Leetcode 刷题笔记1 动态规划part08

关注

原创 2025-03-13 16:09:45 270 阅读 AI 写同款· GEO 优化 ›

Day41 动态规划part08 1.这道题和上一题的最大区别就是dp[i][0] = Math.max(dp[i-1][0], dp[i-1][1] - prices[i]);分析持有股票的情况，如果之前没有持有股票，然后再买入股票，这种情况需要考虑因为之前的买卖操作，此时手头的现金可能就不再是0了，可能会有之前买卖的利润即：dp[i - 1][1]，所以dp[i - 1][1] - prices[i]。对dp[i][0]和dp[i][1]的更新就像121题，然后对dp[i][2]和dp[i][3]的更新就是122题。阅读详情

leetcode 121 买卖股票的最佳时机

把股票问题理解为不卖和卖的两种情况，就只需要考虑两个变量即可

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

leetcode 122 买卖股票的最佳时机 ||

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

上题代码改编：

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

leetcode 123 买卖股票的最佳时机 |||

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        if len(prices) == 0:
            return 0
        length = len(prices)
        dp = [0] * 5
        dp[1] = dp[3] = float('-inf')
        for i in range(length):
            dp[1] = max(dp[1], dp[0] - prices[i])
            dp[2] = max(dp[2], dp[1] + prices[i])
            dp[3] = max(dp[3], dp[2] - prices[i])
            dp[4] = max(dp[4], dp[3] + prices[i])
        return dp[4]

收录于