Leetcode:121. Best Time to Buy and Sell Stock

最新推荐文章于 2025-04-18 22:24:21 发布

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本文介绍了一种寻找股票交易中最大利润的算法。该算法通过一次遍历价格数组，跟踪最低买入价格并计算可能的最大利润。示例展示了如何处理不同情况以确保卖出价格高于买入价格。

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.
本题关键是要先买后卖，而且买的价格一定要比卖的少，否则输出为0

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