【每日一题Leedcode】121. Best Time to Buy and Sell Stock

最新推荐文章于 2021-06-14 18:18:13 发布

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本文介绍了一种寻找股票买卖最佳时机以获得最大利润的算法。通过记录历史最低价格和使用Kadane算法两种方法，文章提供了清晰的实现思路。示例展示了如何在不同情况下计算最大利润。

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

If you were only permitted to complete at most one transaction (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Note that you cannot sell a stock before you buy one.

Example 1:

Input: [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
             Not 7-1 = 6, as selling price needs to be larger than buying price.

Example 2:

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

思路一：记录历史最小价格

class Solution:
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        if len(prices) == 0:
            return 0
        
        Maxprice =  0
        minprice = prices[0]
        for i in range(1,len(prices)): 
            minprice = min(minprice,prices[i])            
            Maxprice = max(prices[i]-minprice,Maxprice)
                
        return Maxprice

思路二：同53题最大子串，利用kadane 算法。

        Maxprice = temp = 0
        for i in range(1,len(prices)):
            temp += prices[i] - prices[i-1]
            temp = max(0,temp)
            Maxprice = max(temp,Maxprice)

参考：https://leetcode.com/problems/best-time-to-buy-and-sell-stock/discuss/39038/Kadane's-Algorithm-Since-no-one-has-mentioned-about-this-so-far-:)-(In-case-if-interviewer-twists-the-input)