题目描述：

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.

原始代码略显麻烦，主要思路是不停改变最大与最小值:

int maxProfit(vector<int>& prices) {
        if(prices.size()==0)
            return 0;
        int max = 0;
        int buyin = prices[0];
        int buyout = prices[0];
        for(int i = 1;i<prices.size();i++){
            if(prices[i]<buyin)
                buyin = prices[i];
            else{
                buyout = prices[i];
                if(buyout - buyin>max)
                    max = buyout -buyin;
            }
        }
        return max;
    }

改进代码简洁明了，但思路相同：

int maxProfit(vector<int> &prices) {
    int maxPro = 0;
    int minPrice = INT_MAX;
    for(int i = 0; i < prices.size(); i++){
        minPrice = min(minPrice, prices[i]);
        maxPro = max(maxPro, prices[i] - minPrice);
    }
    return maxPro;
}