leetcode 121. Best Time to Buy and Sell Stock

蓝羽飞鸟

于 2019-06-15 19:01:19 发布

阅读量81

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode 文章标签：算法

本文链接：https://blog.youkuaiyun.com/level_code/article/details/92152266

leetcode 专栏收录该内容

490 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数组时的最大交易利润。通过一次遍历，更新最低购买价格和最大利润，确保了高效求解。适用于只允许进行一次买卖的情况。

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 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.

数组是每天的股票价格，问在最低价买入最高价卖出的收益，注意卖出的时间要在买入的后面

思路：
顺序访问数组，更新最低价，同时更新最大收益(price[i] - 最低价）

    public int maxProfit(int[] prices) {
        if(prices == null || prices.length == 0) {
            return 0;
        }
        
        int minV = prices[0];
        int result = 0;
        
        for(int i = 1; i < prices.length; i++) {
            result = Math.max(result, prices[i] - minV);
            minV = Math.min(minV, prices[i]);
        }
        
        return result;
    }