Say you have an array for which the ith 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>记录最大最小值的方法。
class Solution(object):
def maxProfit(self, prices):
"""
:type prices: List[int]
:rtype: int
"""
if len(prices)==0:
return 0
profit = 0
pro =[0]
min = prices[0]
for price in prices[1:]:
if price < min:
min = price
if price > min:
profit = price - min
pro.append(profit)
return max(pro)

本文介绍了一种寻找股票买卖最佳时机以实现最大利润的算法。该算法通过一次遍历记录最低买入价格,并计算每次可能的卖出利润,最终返回最大利润。示例中展示了如何在给定的价格序列中找到最佳买卖时机。
753

被折叠的 条评论
为什么被折叠?



