面试经典150题——Day7

文章目录

一、题目

121. Best Time to Buy and Sell Stock

You are given an array prices where pricesi 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.

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 <= pricesi <= 104

题目来源:leetcode

二、题解

使用动态规划的思想,避免双重for循环,在O(n)时间内解决问题

cpp 复制代码
class Solution {
public:
    int max(int a,int b){
        if(a > b) return a;
        else return b;
    }
    int maxProfit(vector<int>& prices) {
        const int N = 100010;
        int n = prices.size();
        int dp[N];
        dp[0] = 0;
        int minVal = prices[0];
        for(int i = 1;i < n;i++){
            if(prices[i] < minVal) minVal = prices[i];
            dp[i] = max(prices[i] - minVal,dp[i-1]);
        }
        return dp[n-1];
    }
};
相关推荐
kisshyshy9 小时前
🍦 雪糕、食堂、火车厢:三幅漫画吃透栈、队列与链表
javascript·算法
众少成多积小致巨11 小时前
JNI (Java Native Interface) 技术手册中文参考指南
android·java·c++
猿人谷17 小时前
不只是 CPU 阈值:STAR 如何用 GAT + Transformer 做容器级自动扩缩容?
人工智能·算法
复杂网络18 小时前
Stable Diffusion 视觉大模型微调技术深度调研
算法
复杂网络18 小时前
基于 Stable Diffusion 架构的视觉大模型代表性工作与原理深度解析
算法
MrZhao40018 小时前
Agent Loop 如何用 Hook 扩展:权限、日志与工具拦截
算法
MrZhao40018 小时前
Agent 为什么需要 Skills:别把所有知识都塞进 system prompt
算法
JieE2122 天前
LeetCode 101. 对称二叉树|JS 递归 + 迭代双解法,彻底搞懂镜像判断
javascript·算法
JieE2123 天前
LeetCode 56. 合并区间|超清晰 JS 图解思路,面试高频区间题
javascript·算法·面试
Jack203 天前
HarmonyOS开发中错误处理策略:网络异常统一处理
算法