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).
Example
Given an example [2,1,2,0,1], return 2
LeetCode上的原题,请参见我之前的博客Best Time to Buy and Sell Stock II。
class Solution { public: /** * @param prices: Given an integer array * @return: Maximum profit */ int maxProfit(vector<int> &prices) { int res = 0, n = prices.size(); for (int i = 0; i < n - 1; ++i) { if (prices[i] < prices[i + 1]) { res += prices[i + 1] - prices[i]; } } return res; } };
本文转自博客园Grandyang的博客,原文链接:买股票的最佳时间之二[LintCode] Best Time to Buy and Sell Stock II ,如需转载请自行联系原博主。