Leetcode Best Time to Buy and Sell Stock

最新推荐文章于 2025-08-08 11:49:43 发布

原创最新推荐文章于 2025-08-08 11:49:43 发布 · 155 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode

leetcode 专栏收录该内容

124 篇文章

订阅专栏

本文介绍了一种算法，用于计算给定股票价格序列中能够获得的最大买卖利润。该算法通过一次遍历找到最大利润，适用于只允许进行一次交易的情况。

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 i^th element is the price of a given stock on day i.

If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Example 1:

Input: [7, 1, 5, 3, 6, 4]
Output: 5

max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)

Example 2:

Input: [7, 6, 4, 3, 1]
Output: 0

In this case, no transaction is done, i.e. max profit = 0.

将其转换为求区间的序列和最大。代码如下：

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if(prices.empty())
                return 0;
        
        vector<int> temp(prices.size()-1);
        int max = 0,sum=0;

        for(int i=0;i<temp.size();i++)
                temp[i] = prices[i+1]-prices[i];
        
        for(int i=0;i<temp.size();i++)
        {
            sum += temp[i];
            if(sum > max)
                max = sum;
            
            if(sum < 0)
                sum = 0;
        }
        return max;
    }
};