leetcode Best Time to Buy and Sell Stock II

Best Time to Buy and Sell Stock II

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).

下面的程式一定要加上if(price.empty()) return 0;才能通過。

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;
	}
           

好難發現的一個bug。size()預設傳回的是無符号整數。

下面程式就不用增加判斷：

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;
	}