121. Best Time to Buy and Sell Stock

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本文介绍了一种算法，用于从给定的股票价格数组中找到能够获得最大利润的买入和卖出时机，仅允许进行一次交易。通过实例展示了如何计算最大利润，并提供了C++实现代码。

problem：

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.

solution：

using namespace std;

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