LeetCode121. 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 · 152 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。该算法通过一次遍历即可找出买进和卖出的最佳时间点，适用于只允许进行一次交易的情况。通过两个示例展示了算法的应用过程。

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

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.

遍历数组
如果当前的价格低于先前确定买入时价格，则将买入时改为当前时间；
如果当前价格高于之前确定卖出的价格，则将卖出时间改为当前时间；
每次和前一步求得的差价做比较去较大值。

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