题目来源
https://leetcode.com/problems/best-time-to-buy-and-sell-stock/description/
题目描述
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.
给定一个数组, 第 i 个值代表了第 i 天的股票价格. 假设你只能做最多一次股票交易, 求出你能赚到的最大利润. (一次交易是指一次买进, 一次卖出, 买进必须在卖出之前.)
解题思路
如果我们把这个价格数组转化为涨跌数组, 就相当于转化成了一个最大子数组求和的问题. 比方说题目中给出的数组是: [7,1,5,3,6,4], 涨跌数组即是[-6,4,-2,3,-2]. (注意涨跌数组肯定比价格数组少一个元素).
很容易看出涨跌数组的最大连续子数组之和便是我们能赚到的最大利润.
代码实现
class Solution:
def maxProfit(self, prices):
"""
:type prices: List[int]
:rtype: int
"""
size = len(prices)
if size <= 1:
return 0
lastMax = prices[1] - prices[0]
ret=max(0,lastMax)
for i in range(1, size - 1):
profit = prices[i + 1] - prices[i]
if lastMax <= 0:
lastMax = profit
else:
lastMax += profit
ret = max(lastMax, ret)
return ret
代码表现
可以看到代码表现还是没上90%略为可惜. 看了看其它表现更好的代码, 发现它们用的策略是:
