有一个价格数组，对应的位置代表改天股票的价格，你有一次买入和卖出的机会，你如何使得收益最大化。也就是返回profit。

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.

其实就是扫一遍数组，记录到当前的最小值，然后当前值减去最小值和profit对比，大的存在profit里面，最后返回profit就行了

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

leetcode Best Time to Buy and Sell Stock