中山大学算法课程题目详解（第七周）

问题描述：

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).

解决思路：

这道题只是对于贪心算法概念的一个简单应用，股票交易次数不限，所以我们对于每一天的股票价格进行遍历，只要发现某天的股票价格大于前一天的价格，这个时候便是盈利的时候，座椅我们把两天股票的差价加到总的盈利中，遍历结束后输出总的盈利即可

具体代码：

int maxProfit(vector<int>& prices) {
	if (prices.size() < 2) {
		return 0;
	}
	int count = 0;
	int current = prices[0];
	for (auto a : prices) {
		if (a > current) {
			count += (a - current);
		}
		current = a;
	}
	return count;
}