题目描述

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

Note that you cannot sell a stock before you buy one.

例子

例子 1

Input: [7,1,5,3,6,4] Output: 5 Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5. Not 7-1 = 6, as selling price needs to be larger than buying price.

例子 2

Input: [7,6,4,3,1] Output: 0 Explanation: In this case, no transaction is done, i.e. max profit = 0.

解题思路

这道题可以用动态规划的思路来做：用一个变量 min_price_before 更新在当前位置之前最低的价格，然后遍历数组，最大化当前卖出去(prices[i] - min_price_before) 能获得的最高利润即可，代码如下：

class Solution {
public:
    int maxProfit(std::vector<int>& prices) {
        if (prices.empty()) return 0;

        int n_days = prices.size();
        int max_profit = 0;

        int min_price_before = prices[0];

        for (int i = 1; i < n_days; i++) {
            max_profit = std::max(max_profit, prices[i] - min_price_before);
            min_price_before = std::min(min_price_before, prices[i]);
        }

        return max_profit;
    }
};

• 时间复杂度: O(n)
• 空间复杂度: O(1)