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).
买卖股票的最好时机，题目要求再卖之前必须先购买股票。所以只需要在遍历价格数组时，碰到相邻元素顺序时求他们的差值后累加即可。
class Solution { public: int maxProfit(vector<int>& prices) { if (prices.size() == 0 || prices.size() == 1) return 0; int res = 0; for (int i = 1; i != prices.size(); i++) if (prices[i - 1] < prices[i]) res += (prices[i] - prices[i - 1]); return res; } }; // 6 ms