Python3解leetcode Best Time to Buy and Sell Stock II

问题描述：

 

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.

 

思路：

只要price还在上涨，那就不能够卖出，上涨价格的累加就是获得的利润；

只有当今天价格比昨天价格低的时候，这个时候才是没有利润，开始新一轮买卖周期的时候

基于以上思想，整个代码可以非常简洁

代码：

 

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        
        return sum( prices[i+1]-prices[i] if(prices[i+1]-prices[i])>0  else 0      for i in range(len(prices)-1)                  )