121. Best Time to Buy and Sell Stock

最新推荐文章于 2024-01-03 12:56:44 发布

t_jianyan

最新推荐文章于 2024-01-03 12:56:44 发布

阅读量333

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode代码笔记

本文链接：https://blog.youkuaiyun.com/t_jianyan/article/details/53206595

leetcode代码笔记专栏收录该内容

82 篇文章

订阅专栏

本文介绍了一种寻找股票交易中最大利润的算法。通过两种不同的方法实现：一是记录每日最低买入价格并计算最大利润；二是利用当前利润与历史最高利润进行比较更新。这两种方法都有效地解决了只允许进行一次买卖的问题。

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 (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Example 1:

Input: [7, 1, 5, 3, 6, 4]
Output: 5

max. difference = 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

In this case, no transaction is done, i.e. max profit = 0.

如果直接用两次迭代，会超时，倒序输入时，会很慢。

DP问题

class Solution {
public:
    int maxProfit(vector<int> &prices) {
    int maxPro = 0;
    int minPrice = INT_MAX;
    for(int i = 0; i < prices.size(); i++){
        minPrice = min(minPrice, prices[i]);
        maxPro = max(maxPro, prices[i] - minPrice);
    }
    return maxPro;
}
};

minPrice is the minimum price from day 0 to day i. And maxPro is the maximum profit we can get from day 0 to day i.

How to get maxPro? Just get the larger one between current maxPro and prices[i] - minPrice.

第二种

class Solution {
public:
    int maxProfit(vector<int> &prices) {
    int maxCur = 0, maxSoFar = 0;
        for(int i = 1; i < prices.size(); i++) {
            maxCur = max(0, maxCur += prices[i] - prices[i-1]);
            maxSoFar = max(maxCur, maxSoFar);
        }
        return maxSoFar;

}
};