122-股票的最大利润（可以多次买卖）_计算机可以买卖三次股票,怎么样收益最大

122股票的最大利润

给定一个数组，它的第 i 个元素是一支给定股票第 i 天的价格。

设计一个算法来计算你所能获取的最大利润。你可以尽可能地完成更多的交易（多次买卖一支股票）。

注意：你不能同时参与多笔交易（你必须在再次购买前出售掉之前的股票）。

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.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17

solution:感觉还是没有用到动态规划的东西，只要后面一天的的股票比前一天高，就把股票的利润加到收益中

class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        # 只要后面的股票价格比前一天高，就把利润添加到收益中。
        maxP = 0
        for i in range(len(prices)-1):
            if (prices[i+1]>prices[i]):
                maxP+=(prices[i+1]-prices[i])
        return maxP
1
2
3
4
5
6
7
8
9
10
11
12

用C++重新写一下：

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if(prices.empty())
            return 0;
        int profit=0;
        for(int i=1;i<prices.size();i++){
            if(prices[i]>prices[i-1])
                profit += prices[i]-prices[i-1];
        }
        return profit;
    }
};
1
2
3
4
5
6
7
8
9
10
11
12
13