leetcode121. Best Time to Buy and Sell Stock

最新推荐文章于 2025-04-18 22:24:21 发布

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最新推荐文章于 2025-04-18 22:24:21 发布

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分类专栏： leetcode

本文链接：https://blog.youkuaiyun.com/orangefly0214/article/details/88066483

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本文介绍了一种通过动态规划算法解决股票交易中寻找最大利润的问题。以数组表示每日股价，目标是在一次买入和卖出中获得最大利润。文章提供了具体实例和代码实现，展示了如何计算最优交易策略。

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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.

解题思路：

用动态规划。本题同剑指offer：连续子序列最大和的思想是一模一样的。https://blog.youkuaiyun.com/orangefly0214/article/details/85628042

实现：

class Solution {
    public int maxProfit(int[] prices) {
        if(prices==null||prices.length==0||prices.length==1){
            return 0;
        }
        int[] diff=new int[prices.length-1];
        for(int i=1;i<prices.length;i++){
            diff[i-1]=prices[i]-prices[i-1];
        }
        int max=diff[0];
        int currSum=diff[0];
        for(int i=1;i<diff.length;i++){
            currSum=Math.max(currSum+diff[i],diff[i]);
            max=Math.max(currSum,max);
        }
        return max>0?max:0;
    }
}