Best Time to Buy and Sell Stock II

最新推荐文章于 2022-11-16 09:46:40 发布

原创最新推荐文章于 2022-11-16 09:46:40 发布 · 367 阅读

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本文介绍了一种算法，用于计算给定股票价格序列时能够获得的最大利润。该算法允许进行多次买卖交易，但规定每次买入前必须先卖出。通过寻找所有递增序列并计算其差值之和来确定最大收益。

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 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).

pro:给一个数组，第i个数字是第i天的股票价格，现在能完成任意多次的交易(一次交易指买进和卖出一股)，但是一天之内不能涉及多次交易，也就是必须在买进之前卖出，求最大获利

sol:也就是求这个数组中某些相离的start，end，和是最大获利。

find all the increasing sequence,result is the sum of (end-start) of all sequences.
because in the middle of sequence,there is no point to cut it.the biggest profit is end-start

code:

class Solution
{
public:
	int maxProfit(vector<int> &prices)
	{
		int i,j,len;
		if(prices.size()<=1)  return 0;
		int res=0;
		int minn,maxx;
		minn=maxx=prices[0];
		for(i=1;i<prices.size();)
		{
			//cout<<"***"<<i<<' '<<prices[i]<<endl;
			while(i<prices.size()&&prices[i]>prices[i-1])
			{
				maxx=prices[i];
				i++;
			}
			res+=maxx-minn;
			if(i<prices.size())
			{
				minn=maxx=prices[i];
				i++;
				while(i<prices.size()&&prices[i]<prices[i-1])
				{
					minn=maxx=prices[i];
					i++;
				}
			}
		}
		return res;
	}
};