leetcode-买卖股票的最佳时机④*

题目描述：

方法一：

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