[Array]122. Best Time to Buy and Sell Stock II

• 分类：Array/Greedy
• 时间复杂度: O(n)
• 空间复杂度: O(1)

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.

代码：

自己想的代码：

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        # edge
        if prices==None or len(prices)<2:
            return 0
        
        profit=0
        buyprice=prices[0]
        i=0
        while i=prices[i-1]:
                    i+=1
                i-=1
                profit+=(prices[i]-buyprice)
            i+=1
        return profit

GTH代码思路：

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        # edge
        if prices==None or len(prices)<2:
            return 0
        
        profit=0
        for i in range(1,len(prices)):
            if prices[i]>prices[i-1]:
                profit+=(prices[i]-prices[i-1])

        return profit

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讨论：

1.GTH的思路好简洁好聪明好干净啊
2.不是我聪明，而是人家代码写的好我看懂了...