[Leetcode] 121. Best Time to Buy and Sell Stock 解题报告

最新推荐文章于 2025-08-19 21:22:03 发布

原创最新推荐文章于 2025-08-19 21:22:03 发布 · 437 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#Leetcode #解题报告 #Array #Greedy

IT公司面试习题专栏收录该内容

736 篇文章

订阅专栏

本文介绍了一种简单高效的算法来解决股票交易中寻找最佳买卖时机的问题。通过一次遍历即可找到最大利润，时间复杂度为O(n)，空间复杂度为O(1)。

题目：

Say you have an array for which the ith 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.

思路：

Easy级别的题目。我们记录一个截止当前日期的最低买入价，以及一个截止当前日期的最大获利，然后线性扫描数组，一旦发现当前卖出的获利大于最大获利值，则更新最大获利值；如果当前买入价小于最低买入价，则更新最低买入价。算法的时间复杂度是O(n)，空间复杂度是O(1)。

代码：

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if (prices.size() == 0) {
            return 0;
        }
        int lowest_price = prices[0];
        int max_profit = 0;
        for (int i = 1; i < prices.size(); ++i) {
            if (prices[i] - lowest_price > max_profit) {
                max_profit = prices[i] - lowest_price;
            }
            if (prices[i] < lowest_price) {
                lowest_price =prices[i];
            }
        }
        return max_profit;
    }
};