买卖股票的最佳时机 II

最新推荐文章于 2023-03-20 17:06:14 发布

JustForYouForNLP

最新推荐文章于 2023-03-20 17:06:14 发布

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分类专栏：力扣刷题集锦

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想法是取到低峰和高峰相减，我的解法：

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        
        if not prices:
            return 0
        
       
        benefit = 0
        
        i = 0
        lens = len(prices)
        while i < lens:
            minVal = prices[i]
            maxVal = prices[i]
            j = i
            while j < lens:
                if prices[j] <= minVal:
                    minVal = prices[j]
                    maxVal = prices[j]
                    j += 1
                else:
                    break
                    
            while j < lens:
                if prices[j] >= maxVal:
                    maxVal = prices[j]
                    j += 1
                else:
                    break
            
            benefit += maxVal - minVal
            i = j
        
        return benefit

大佬的解法：简单粗暴，直接取增长值即可

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        
        if not prices:
            return 0
        
        benefit = 0
        
        for i, p in enumerate(prices[1:]):
            if p > prices[i]:
                benefit += p - prices[i]
        
        return benefit