122.md

122. Best Time to Buy and Sell Stock II

难度: Easy

刷题内容

原题连接

• https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/description/

内容描述

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.

解题方案

思路 1 - 时间复杂度: O(N) - 空间复杂度: O(1)

beats 22.56%

因为不限制买卖次数，所以对于除了最后一天的每一天，我们只需要看看它是否比明天价格低即可， 如果是的话那我们就今天买入明天卖出（注意：前后两天价格相等的话我们不做买卖操作，因为可能会减少后面可以赚钱的操作数），这样一直操作下去叠加即可

class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        if not prices or len(prices) == 0:
            return 0
        return sum([max(prices[i+1]-prices[i], 0) for i in range(len(prices)-1)])

如果可读性高点可以这样

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