LeetCode: Best Time to Buy and Sell Stock III

最新推荐文章于 2022-07-03 21:43:22 发布

原创最新推荐文章于 2022-07-03 21:43:22 发布 · 328 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 专栏收录该内容

186 篇文章

订阅专栏

本文介绍了一个寻找最大股票利润的算法，允许进行最多两次交易，且不允许同时进行多笔交易。通过设计一个类`Solution`，实现算法并提供详细步骤，包括初始化变量、迭代计算最大利润。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 at most two transactions.

Note:

You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again)

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        if(prices.empty() || prices.size() == 1)
            return 0;
        int size = prices.size();             
        vector<int> left(size, 0);
        vector<int> right(size, 0);
        int result = 0;
        int min = prices[0];
        int max = prices[size-1];
        for(int i = 1; i < size; i++)
        {
            if(prices[i] < min)
                min = prices[i];
            left[i] = std::max(left[i-1], prices[i] - min);
        }
        for(int i = size-2; i >= 0; i--)
        {
            if(prices[i] > max)
                max = prices[i];
            right[i] = std::max(right[i+1], max - prices[i]);
        }
        for(int i = 0; i < size; i++)
        {
            if(left[i] + right[i] > result)
                result = left[i] + right[i];
        }
        return result;
    }
};

Round 2:

class Solution {
public:
    int maxProfit(vector<int>& prices) {
		if(prices.size() == 0)
			return 0;
        int *left = (int *)calloc(prices.size(), sizeof(int));
		int *right = (int *)calloc(prices.size(), sizeof(int));
		int min = prices[0], max = prices[prices.size()-1];
		int result = 0;
		for(int i = 1; i < prices.size(); i++)
		{
			int j = prices.size() - i - 1;
			min = std::min(prices[i], min);
			max = std::max(prices[j], max);
			left[i] = std::max(left[i-1], prices[i] - min);
			right[j] = std::max(right[j+1], max - prices[j]);
		}
		for(int i = 0; i < prices.size(); i++)
		{
			if(left[i] + right[i] > result)
				result = left[i] + right[i];
		}
		return result;
    }
};