leetcode122 Best Time to Buy and Sell Stock II

"""
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.
"""
"""
搞笑的一题，容易想复杂
可以与leetcode121联系来看https://www.cnblogs.com/yawenw/p/12261726.html
其实这个题要求一天不能同时买和卖股票，
而且必须再一笔交易完成之后才能进行下一笔交易
假设序列是a <= b <= c <= d  [1, 2, 3, 4]
最大利润 d - a = (b - a) + (c - b) + (d - c)
假设序列是a <= b， b >= c, c <= d [1, 5, 4, 8]
最大利润(b - a) + (d - b)
"""
class Solution:
    def maxProfit(self, prices):
        res = 0
        for i in range(len(prices)-1):
            if prices[i] < prices[i+1]:
                res += prices[i+1]-prices[i]
        return res