1 """
2 Say you have an array for which the ith element is the price of a given stock on day i.
3 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).
4 Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).
5 Example 1:
6 Input: [7,1,5,3,6,4]
7 Output: 7
8 Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
9 Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
10 Example 2:
11 Input: [1,2,3,4,5]
12 Output: 4
13 Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
14 Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
15 engaging multiple transactions at the same time. You must sell before buying again.
16 Example 3:
17 Input: [7,6,4,3,1]
18 Output: 0
19 Explanation: In this case, no transaction is done, i.e. max profit = 0.
20 """
21 """
22 搞笑的一题，容易想复杂
23 可以与leetcode121联系来看https://www.cnblogs.com/yawenw/p/12261726.html
24 其实这个题要求一天不能同时买和卖股票，
25 而且必须再一笔交易完成之后才能进行下一笔交易
26 假设序列是a <= b <= c <= d [1, 2, 3, 4]
27 最大利润 d - a = (b - a) + (c - b) + (d - c)
28 假设序列是a <= b， b >= c, c <= d [1, 5, 4, 8]
29 最大利润(b - a) + (d - b)
30 """
31 class Solution:
32 def maxProfit(self, prices):
33 res = 0
34 for i in range(len(prices)-1):
35 if prices[i] < prices[i+1]:
36 res += prices[i+1]-prices[i]
37 return res
