leetcode Best Time to Buy and Sell Stock II

最新推荐文章于 2022-07-03 21:43:22 发布

最新推荐文章于 2022-07-03 21:43:22 发布 · 75 阅读

本文介绍了一种在股票市场中利用价格波动获取最大利润的算法。通过遍历价格数组并只在价格上涨时进行买入和卖出操作，实现多次交易以最大化收益。程序演示了如何编写有效算法来应对不同市场条件，从而提高投资者的决策效率。

Best Time to Buy and Sell Stock II

Say you have an array for which theithelement is the price of a given stock on dayi.

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).

下面的程序一定要加上if(price.empty()) return 0;才能通过。

int maxProfit(vector<int> &prices) 
	{
		if (prices.empty()) return 0;
		int profit = 0;
		for (int i = 0; i < prices.size()-1; i++)
		{
			if (prices[i] < prices[i+1]) profit += prices[i+1]-prices[i];
		}
		return profit;
	}

也可以修改如下：

int maxProfit(vector<int> &prices) 
	{
		//if (prices.empty()) return 0;
		int profit = 0;
		for (int i = 0; i < int(prices.size())-1; i++)
		{
			if (prices[i] < prices[i+1]) profit += prices[i+1]-prices[i];
		}
		return profit;
	}

好难发现的一个bug。size()默认返回的是无符号整数。

下面程序就不用增加判断：

int maxProfit2(vector<int> &prices) 
	{
		int profit = 0;
		for (int i = 1; i < prices.size(); i++)
		{
			if (prices[i-1] < prices[i]) profit += prices[i]-prices[i-1];
		}
		return profit;
	}

//2014-2-17 update
	int maxProfit(vector<int> &prices) 
	{
		int max_profit = 0;
		for (int i = 1; i < prices.size(); i++)
		{
			if (prices[i] > prices[i-1]) max_profit += prices[i]-prices[i-1];
		}
		return max_profit;
	}