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

1:遍历数组；2:每个数字获得该数字之前的最小的数字，并与当时保存的最大值相比較

    int maxProfit(vector<int> &prices)

    {

        if(prices.size() <= 1)

        {

            return 0;

        }

        int maxValue = 0;

        int minPrice = prices[0];

        int size = (int)prices.size();

        for(int i = 1; i < size; i++)

        {

            if(prices[i] > minPrice)

            {

                maxValue = (maxValue > prices[i] - minPrice ? maxValue : prices[i] - minPrice);

            }

            else

            {

                minPrice = prices[i];

            }

        }

        return maxValue;

    }