目录

学习目标

学习内容

 121. 买卖股票的最佳时机 

 122.买卖股票的最佳时机II 

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学习目标

学习内容

 121. 买卖股票的最佳时机 

121. 买卖股票的最佳时机 - 力扣（LeetCode）https://leetcode.cn/problems/best-time-to-buy-and-sell-stock/

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

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

 122.买卖股票的最佳时机II 

122. 买卖股票的最佳时机 II - 力扣（LeetCode）https://leetcode.cn/problems/best-time-to-buy-and-sell-stock-ii/

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

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