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 · 69 阅读

本文介绍了一种无限次股票买卖策略，旨在通过多次买入卖出实现最大利润，且不同时进行多笔交易。提供了三种不同的算法实现，包括判断空数组、使用循环遍历价格数组并计算利润的方法。最后更新了代码，优化了性能。

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;
	}