122. Best Time to Buy and Sell Stock II 超时解决

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最新推荐文章于 2024-12-31 20:00:00 发布

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本文介绍了一种设计算法，用于计算给定股票价格数组中可能获得的最大利润。算法允许进行多次买卖交易，但每次卖出后才能再次购买。通过示例说明了如何在不同价格走势下计算最大利润。

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Say you have an array for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete as many transactions as you like (i.e., buy one and sell one share of the stock multiple times).

Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).

Example 1:

Input: [7,1,5,3,6,4]
Output: 7
Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
             Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.

Example 2:

Input: [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
             Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
             engaging multiple transactions at the same time. You must sell before buying again.

Example 3:

Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

数学模型是：求和（所有的peak值-所有的最低值）

class Solution:
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        
        maxprofit,i=0,0
        while(i < len(prices)-1):
            while(prices[i] >= prices[i+1]) and (i<len(prices)-2):
                i=i+1
            lowest=prices[i]
            while(prices[i]<=prices[i+1]) and (i<len(prices)-2):   
                i=i+1
            highest=prices[i]
            maxprofit += highest - lowest;
        return maxprofit

改成下面的则通过：

数学模型是：求和（所有的peak值-所有的最低值）====> 等价于所有的差值之和

 i,maxprofit=1,0
        while(i<len(prices)):
            diff=prices[i]-prices[i-1];
            if diff > 0:
                maxprofit =maxprofit + diff
            i=i+1
        return maxprofit