Best Time to Buy and Sell Stock II

Say you have an array for which the *i*th 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.

这是昨天刷的 Best Time to Buy and Sell Stock 进阶，允许多次买入卖出，求最大利润

如果允许多次交易，如何保证最大利益，肯定是每个低点买下个高点卖，所以第一种解法就是求出每个谷峰和谷底

如何求谷底，向前比较只要比前一个大就说明还在降低，只要停止降低，下一个就是最低价，谷峰相反

class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        i = 0
        vallley = prices[0] if prices else 0
        peak = prices[0] if prices else 0
        profit = 0
        
        # 这里的 -1 很重要 因为最后以为再向前走会超出范围，每次都会向前比较一位 所以可以跳过 很重要的细节
        while i < (len(prices) - 1):
            # 找出谷底
            while i < (len(prices) - 1) and prices[i] >= prices[i+1]:
                i += 1
            valley = prices[i]
            # 找出谷峰
            while i < (len(prices) - 1) and prices[i] <= prices[i+1]:
                i += 1
            peak = prices[i]
            profit += peak - valley
        
        return profit

还有一种思路，可以只找增长的部分，不用找谷峰谷底，每次向后比较，只要比上一次高就算入利润中

class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        profit = 0
        for i in range(len(prices)):
            if i > 0 and prices[i] > prices[i-1]:
                profit += prices[i] - prices[i-1]
        
        return profit

这种思路只用关心增长就好了，代码很简约直白。

发现自己脑中的想法总是很浅，没有形成一种体系，还是深度思考比较少，对一些问题缺少体系化的认知，缺少体系化的阅读，可能是碎片化阅读带来的结果，以后需要注意了，遇到事情多问几个为什么，不能一味接受。

叠加式的进步，或者原地踏步。