121. Best Time to Buy and Sell Stock

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博客围绕股票交易问题，要求在最多完成一次交易的情况下找到最大利润。给出了两种算法代码实现，代码1通过遍历数组记录最小价格来计算最大利润，代码2使用双重循环，时间复杂度较高，还展示了两者的运行时间。

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.

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.

代码1：

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int size = prices.size();
        int out = 0;
        if(size == 0 || size == 1)
            return out;
        int minPrice = prices[0];
        for(int i = 1; i<size; i++){
            minPrice = min(minPrice, prices[i]);
            out = max(out, prices[i]-minPrice);
        }
        return out;
    }
};

代码2：（这里只是对比代码1，代码2时间复杂度太高）

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int out =0;
        int price = 0;
        int priceadd = 0;
        int size = prices.size();
        if(size == 0 || size == 1)
            return out;
        for(int i = 0; i<size-1; i++){
            price = prices[i];
            for(int j = i+1; j<size; j++){
                if(prices[j]>price){
                    priceadd = prices[j]-price;
                    out = out < priceadd ? priceadd : out;
                }
            }
        }
        return out;
    }
};