122. Best Time to Buy and Sell Stock II（python+cpp）

题目：

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.

解释：
一共可以进行任意多次操作，但是在买入一个stock之前必须先卖出手中已有的stock，每天可以先卖出已有的stock再买入新的stock，求能获得的最大利润。
python代码：

class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        return sum([max(prices[i+1]-prices[i],0) for i in xrange(len(prices)-1)])

c++代码：

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if(prices.size()==0)
            return 0;
        int _sum=0;
        for (int i=0;i<prices.size()-1;i++)
        {
            _sum+=max(0,prices[i+1]-prices[i]);
        }
        return _sum;
    }
};

总结：
注意题目所给条件。