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).
这个问题首先要理解这是一只股票,你可以今天买,在今后的某一天卖,然后再买回,再卖。分析一下实际上就是一个简单的求和maxProfit = ∑max(prices[i + 1] - prices[i], 0)
class Solution { public: int maxProfit(vector<int> &prices) { // IMPORTANT: Please reset any member data you declared, as // the same Solution instance will be reused for each test case. if(prices.size() == 0) return 0; int max_profit = 0; for(int i = 1;i < prices.size();i++) max_profit += max(prices[i] - prices[i - 1], 0); return max_profit; } };