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 (ie, buy one and sell one share of the stock multiple times). However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

解释题意：一个数组储存着一只股票的价格走势，第i个元素就可以看做是该股票第i天的价格，你可以买卖股票，无需考虑手上的钱够买多少股，题目意思是可以随便买卖，但要买就买一股，不卖出这一股是无法再次买入的。这题很简单，第二天比今天价格高了就买入，低了就卖出即可

思路：遍历数组，计算所有第二天比前一天价格高的值，加起来就是最后的最大收益。

class Solution:
    # @param prices, a list of integer
    # @return an integer
    def maxProfit(self, prices):
        profit = 0
        length = len(prices)
        for i in range(0,length-1):
            if prices[i+1] > prices[i]:
                profit += prices[i+1] - prices[i]
        return profit