[LeetCode] 122. 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.

这道题优化的解法是O(N)，想法比较简单，就是求所有波峰波谷之间差的和。只要后一天的股票价格比当前的值高，就把这个价差加到profit上面。如果第二天的值更小，就是一个新的波峰波谷，一切重来。

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        if not prices: return 0
        buy, profit = 0, 0
        for i in range(len(prices) - 1):
            if prices[i] < prices[i+1]:
                profit += prices[i+1] - prices[i]
        return profit