題目鏈接
tag:

• Easy；

question：
??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: you cannot sell a stock before you buy one.

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.

思路：
??比較簡單，只需要遍歷一次數組，用一個變量記錄遍歷過程中的買入股票最小值，然后每次計算當前值和這個最小值之間的差值為利潤，然后每次選較大的利潤來更新。當遍歷完成即為最大利潤，代碼如下：

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int res=0, buy=INT_MAX;
        for(int price : prices) {
            buy = min(buy, price);
            res = max(res, price-buy);
        }
        return res;
    }
};