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 (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).

提示：

此题应使用贪心算法的思想。我所采用的贪心规则如下：

从第一个元素开始遍历至最后一个元素；
若下一个元素比当前元素大，那么就抛售；并以下一个元素的价钱买入，同时累加利润；
若下一个元素比当前元素小，那么就以下一个元素的价钱买入（若连续出现递减的情况，那么最后的买入价格是最后那个最小的元素）；

代码：

class Solution {

public:

    int maxProfit(vector<int>& prices) {

        if (prices.size() == ) return ;

        int buy_in = prices[], sum = ;

        for (int i = ; i < prices.size(); ++i) {

            if (prices[i] > buy_in) {

                sum += prices[i] - buy_in;

                buy_in = prices[i];

            } else {

                buy_in = prices[i];

            }

        }

        return sum;

    }

};

后来在论坛上发现了更加简洁的解法，核心思想其实是比较类似的：

int maxProfit(vector<int> &prices) {

    int ret = ;

    for (size_t p = ; p < prices.size(); ++p)

      ret += max(prices[p] - prices[p - ], );

    return ret;

}

其实就是只要有涨就先卖再买，但是这种解法更加简洁清楚。

巴特西

【LeetCode】122. Best Time to Buy and Sell Stock II

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