[LeetCode]121. Best Time to Buy and Sell Stock（股票买卖的最佳时间）

最新推荐文章于 2020-07-30 16:02:50 发布

原创最新推荐文章于 2020-07-30 16:02:50 发布 · 186 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode

LeetCode 专栏收录该内容

16 篇文章

订阅专栏

本文介绍了一种寻找股票买卖最佳时机的算法。给定一系列每日股票价格，算法的目标是在一次买卖操作中实现最大利润。通过记录遍历过程中的最低购买价格，并计算当前价格与最低购买价格的差值，从而找到最大利润。

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.

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

Note that you cannot sell a stock before you buy one.

Example 1:

Input: [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 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
Explanation: In this case, no transaction is done, i.e. max profit = 0.

题意解读：给定一个数组，数组的第i个元素是第i天的股票价格。如果你只允许进行一次买卖操作（买和买当做一次交易），设计一个算法寻找最大额收益
思路：

1、用一个变量来记录遍历过的数组中的最小值，然后计算当前值和最小值的差值，然后每次选择最大利润来更新。当遍历数组之后，当前利润即为所求利润。

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int res=0,buy=INT_MAX;
        for(auto price :prices)
        {
            buy=min(price,buy);
                res=max(res,price-buy);
        }
        return res;
    }
};