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).
1 class Solution { 2 public: 3 int maxProfit(vector<int>& prices) { 4 if(prices.empty()) 5 return 0; 6 int price = prices[0]; 7 int profit=0; 8 for(auto it = prices.begin();it!=prices.end();++it) 9 { 10 if(price<*it) 11 profit += *it - price; 12 price = *it; 13 } 14 return profit; 15 } 16 };
和最大子序列思路差不多。这里是前一天和后一天比较，如果前一天价格小于后一天就先买后卖，否则这一天不交易。