Best Time to Buy and Sell Stock

最新推荐文章于 2025-04-18 22:24:21 发布

原创最新推荐文章于 2025-04-18 22:24:21 发布 · 146 阅读

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本文介绍了一种寻找股票买卖最佳时机以获得最大利润的算法。输入为一系列每日股价，目标是在仅能进行一次买卖的情况下找到最大收益。文章通过示例说明了如何避免因卖价低于买价而导致的错误计算，并提供了一个初始的暴力解决方法。

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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 (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.

开始使用的是暴力方法，列举出所有可能的交易情况。时间复杂度是O(n^2)。结果超时了。class Solution {
public:
    int maxProfit(vector<int>& prices) {
            int pro=0;
            int sub=0;//sub暂存所有满足条件的利润，从中选择最大利润
            for(int i=0;i<prices.size()-1;i++)//i买入的价格 
              for(int j=i+1;j<prices.size();j++)//j卖出的价格
              {
                  if(prices[i]<prices[j])//且买入在前，卖出在后
                  {
                      sub=prices[j]-prices[i];
                      pro=max(pro,sub);
                  }

              }
        return pro;

    }
};