Best Time to Buy and Sell Stock III

Say you have an array for which the i^th element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete at most two transactions.

此題看清題，最多倆次交易。且倆次交易不能重疊。

個人第一次思路，計算f(0,i)與f(i,n)的最大值。可知O(N^2) 代碼超時！但正確 。

int maxProfit(vector<int> &prices) 
{

	if(prices.size() <2 )return 0;
	int max_profit = 0;
	int min_val ;
	for(int k = 0;k < prices.size();k++) 
	{

		min_val = prices[0];
		int profit_1 = 0;

		for(int i = 1;i < k;i++)
		{
			if(min_val > prices[i])
			{
				min_val = prices[i];
			}
			if(profit_1 < prices[i] - min_val)
			{
				profit_1 = prices[i] - min_val;
			}

		}

		min_val = prices[k];
		int profit_2 = 0;

		for(int i = k+1;i < prices.size();i++)
		{
			if(min_val > prices[i])
			{
				min_val = prices[i];
			}
			if(profit_2 < prices[i] - min_val)
			{
				profit_2 = prices[i] - min_val;
			}

		}
		int val = profit_2 + profit_1;
		if(max_profit < val)
		{
			max_profit = val;
		}
	}

	return max_profit;
}

第二次：感覺，需要緩存。參看他人的代碼！http://www.cnblogs.com/lihaozy/archive/2012/12/19/2825525.html
利用緩存，分別0--》n 計算最大利潤值。再重n--->0計算最大利潤值。

最後利用緩存值，找倆次最大和！O(n)

#include <iostream>
#include <vector>
using namespace std;

int maxProfit(vector<int> &prices);

int main()
{
	int a[] = {2,1,2,0,1};
	vector<int> prices(a,a+5);
	cout<<maxProfit(prices)<<endl;

	return 0;
}

int maxProfit(vector<int> &prices) 
{

	if(prices.size() <2 )return 0;

	int max_profit = 0;
	int min_val ;
	int max_val;

	vector<int> profit_left(prices.size());
	vector<int> profit_right(prices.size());

	profit_left[0] = 0;
	profit_right[prices.size()-1] = 0;

	min_val = prices[0];
	int max_profit_left = 0;
	for(int i = 1;i < prices.size();i++) 
	{
		if(min_val > prices[i])
		{
			min_val = prices[i];
		}
		if(max_profit_left < prices[i] - min_val)
		{
			max_profit_left = prices[i] - min_val;
		}
		
		profit_left[i] = max_profit_left;
		
	}

	max_val = prices[prices.size()-1];
	int max_profit_right = 0;
	for(int i = prices.size()-2;i >= 0;i--) 
	{    

		if(max_val < prices[i])
		{
			max_val = prices[i];
		}
		if(max_profit_right < max_val -prices[i] )
		{
			max_profit_right = max_val -prices[i];
			 
		}
		
		profit_right[i] = max_profit_right;
		

	}

	for(int i = 0; i < prices.size();i++)
	{
		cout<<profit_left[i] <<" "<< profit_right[i]<<endl; 
		if(max_profit < profit_left[i] + profit_right[i])
		{
			max_profit = profit_left[i] + profit_right[i];
		}
	}

	return max_profit;
}