Best Time to buy && sell stock

最新推荐文章于 2025-04-18 22:24:21 发布

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最新推荐文章于 2025-04-18 22:24:21 发布

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分类专栏： LeetCode 跳槽

本文链接：https://blog.youkuaiyun.com/Tiaocaojiayou/article/details/116442054

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该博客讨论了如何在给定股票价格数组中找到最佳的买卖时机以最大化利润。作者使用动态规划策略，分别从左右两端填充两个数组，记录在每个点进行交易可能获得的最大利润，最终通过遍历找到最多两次交易的最大收益。文章还涵盖了特殊情况的处理，例如所有价格相同的情况。

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买卖一次

1个变量记录目前已知最便宜，（当前 - 已知最便宜）= 当前最大收益

买卖无数次 https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii

特殊值的处理，如果全部一样的话，其实可以直接initiate peak && valley = prices[0]，这样对于总体的profit是没有变化的～

t1 < t2 < t3:

buy t1, sell t2 && buy t2 && t3 = buuy t1 && sell t2

这就是得到valley && peak方法的逻辑

profit = Sum (peak - valley)

最多两次

分别用array来记录按照中间某个断点，左右分别可能的最大profit，有一种edge case是只需要一次交易，所以需要最后再多扫描一次

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        
        # Edge case
        if prices is None or len(prices) <= 1:
            return 0
        
        # Divide && Counter
        left_profits = [0] * len(prices)
        left_profits[0] = 0
        
        right_profits = [0] * len(prices)
        right_profits[len(prices) - 1] = 0
        
        # Populate left_profits
        i = 1
        low = prices[0]
        profit = 0
        while (i < len(prices)):
            if prices[i] - low > profit:
                profit = prices[i] - low
            left_profits[i] = profit
            
            # Update knwon low price
            if prices[i] < low:
                low = prices[i]
            
            i = i + 1
            
        # Populate right_profits
        i = len(prices) - 2
        hi = prices[i + 1]
        profit = 0
        while i >= 0:
            if hi - prices[i] > profit:
                profit = hi - prices[i]
            right_profits[i] = profit
            
            # Update known hi price
            if prices[i] > hi:
                hi = prices[i]
            
            i = i - 1
        
        # 3rd pass && get what is the overall best price
        ret_profit = 0
        
        # If has to conduct 2 transactions
        for i in range(0, len(prices) - 1):
            ret_profit = max(ret_profit, left_profits[i] + right_profits[i + 1])
        
        # Edge case is it might just need 1 transaction
        for i in range(0, len(prices)):
            ret_profit = max(ret_profit, max(left_profits[i], right_profits[i]))
        
        return ret_profit