[Leetcode 123, Hard] Best Time to Buy and Sell Stock III

最新推荐文章于 2021-06-15 00:27:24 发布

simplify2012

最新推荐文章于 2021-06-15 00:27:24 发布

阅读量417

点赞数

CC 4.0 BY-SA版权

分类专栏： Leetcode_hard 文章标签： leetcode C++

本文链接：https://blog.youkuaiyun.com/u012212811/article/details/47035293

Leetcode_hard 专栏收录该内容

25 篇文章

订阅专栏

本文介绍了一种算法，用于计算最多完成两笔交易时的最大利润。通过一次遍历找到从左到右的最大利润，并结合从右到左的最高价格来确定最终结果。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Problem:

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

Analysis:

Solutions:

C++:

    int maxProfit(vector<int> &prices) 
    {
        if(prices.size() <= 1)
            return 0;

        int max_left = 0;
        vector<int> maxs_from_left;
        maxs_from_left.push_back(0);
        int min_price = prices[0];
        for(int i = 1; i < prices.size(); ++i) {
	        if(min_price > min(prices[i - 1], min_price))
	            min_price = prices[i - 1];

	        if(max_left < prices[i] - min_price)
	            max_left = prices[i] - min_price;
	            
	        maxs_from_left.push_back(max_left);
        }

        int max_two_trans = maxs_from_left[maxs_from_left.size() - 1];

        int max_price = prices[prices.size() - 1];
        for(int i = prices.size() - 2; i >= 1; --i) {
	        if(max_price < max(max_price, prices[i])) 
	            max_price = prices[i];
	            
	        if(max_two_trans < maxs_from_left[i - 1] + max_price - prices[i])
	            max_two_trans = maxs_from_left[i - 1] + max_price - prices[i];
        }

        return max_two_trans;
    }

Java:

Python: