Leetcode--Java--121. Best Time to Buy and Sell Stock

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#java #枚举模拟

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这篇博客介绍了如何使用动态规划解决股票交易问题，以获得最大利润。通过枚举卖出股票的日期并维护最小购买价格，可以确定最佳买卖时机，从而计算出最大可能的利润。例如，在给定的股票价格数组中，通过在第2天买入并在第5天卖出，可以获得5的利润。如果无法实现利润，则返回0。文章强调了在实际操作中必须先买后卖的原则，并提供了高效的代码实现。

题目描述

You are given an array prices where prices[i] is the price of a given stock on the ith day.

You want to maximize your profit by choosing a single day to buy one stock and choosing a different day in the future to sell that stock.

Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return 0.

样例描述

Example 1:

Input: prices = [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.
Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.
Example 2:

Input: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transactions are done and the max profit = 0.
 

Constraints:

1 <= prices.length <= 105
0 <= prices[i] <= 104

思路

枚举模拟哪一天卖出，然后不断维护这天卖出的利润，和前面的最小买入的价格
控制一个条件（卖出或者买入的日子），维护两个变量即可
注意如果利润小于0就不买入，可以预先设定res = 0

代码

class Solution {
    public int maxProfit(int[] prices) {
       int res = 0;
       //枚举模拟 哪一天卖出  维护最大利润和前面的最小的卖出的价格
       for (int i = 0, minp = 99999; i < prices.length; i ++){
           res = Math.max(res, prices[i] - minp);
           minp = Math.min(minp, prices[i]);
       }
       return res;
    }
}