LeetCode（121） Best Time to Buy and Sell Stock

最新推荐文章于 2022-07-03 21:43:22 发布

kiey_

最新推荐文章于 2022-07-03 21:43:22 发布

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本文链接：https://blog.youkuaiyun.com/cokkiey/article/details/52975099

本文介绍了一种寻找股票买卖最大利润的算法。该算法通过遍历价格数组并记录每个交易日之前的历史最低价格来确定最佳买入时机，进而计算出最大利润。文章提供了具体的实现代码示例。

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题目：选定某天买入，另一天卖出。只能最多一次买和卖，求最大利润

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.

解法：

假设最优的策略是在第k天卖出，那么买入的肯定是前k-1天的最低的价格。所以遍历一遍数组，对于每一天，都假设是在那一天卖出，利润就是那一天的价格减去前面几天的最小值。如果得到的利润比最大利润大，就更新最大利润值。复杂度O(n).

代码：

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