121. Best Time to Buy and Sell Stock

博客围绕LeetCode上“买卖股票的最佳时机”问题展开。给定数组表示每日股票价格,要求设计算法找出最多进行一次交易的最大利润。介绍了遍历数组更新最小元素值并记录最大差值的解法,且该方法只需遍历一遍,优于暴力解法。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

121. Best Time to Buy and Sell Stock

https://leetcode.com/problems/best-time-to-buy-and-sell-stock/

Description

Say you have an array for which the i t h i^{th} 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.

Solutions

思路:
将第一个元素标为min_element,指针向后移动,遍历整个数组,更新最小的元素值,记住最大那个差值。

Submissions

class Solution {
    public int maxProfit(int[] prices) {
        int minprice = Integer.MAX_VALUE;
        int maxprofit = 0;
        for (int i = 0; i < prices.length; i++) {
            if (prices[i] < minprice)
                minprice = prices[i];
            else if (prices[i] - minprice > maxprofit)
                maxprofit = prices[i] - minprice;
        }
        return maxprofit;
    }
}

Summary

只需要遍历一遍,比brute force要好。

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值