# 【LeetCode】 121. 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 (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.

> 給你一個陣列代表，每個元素代表第i天的物價。

> 如果你只能得到一次的價差(只做一次買和賣)，請設計一個演算法找到最大利潤。

> 注意你不能在買之前就賣一個東西。

## Example:

Example 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.
             
             
Example 2:

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

## Solution

* 記住兩個變數：現階段最低價格和現階段最大利潤。
* ForLoop跑每個元素，先計算最大利潤，再更新最低價格。

### Code

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int min_price=9999;
        int max_profit=0;
        for(int i=0;i<prices.size();i++)
        {
            max_profit = max(prices[i]-min_price,max_profit);
            min_price = min(min_price,prices[i]);
        }
        return max_profit;
    }
};

###### tags: `LeetCode` `C++`