LeetCode 121. Best Time to Buy and Sell Stock

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121. Best Time to Buy and Sell Stock

一、问题描述

Say you have an array for which the i*th 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.

二、输入输出

Input: [7, 1, 5, 3, 6, 4]
Output: 5max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)

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

三、解题思路

• 后面会有这道题的变体版，针对这个Easy版本，只需要双重遍历数组，记录并更新maxProfit即可
• 这里只能买卖1次股票使得问题变的非常简单

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

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