题目： 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.
思路分析： 给定一个数组，数组中第i个元素代表第i天股票的价格。通过一次交易（即一次买入和一次卖出）获得最大的收益。 因为我们要以低价买入，以高价卖出才能获得最大收益。所以这道题目转换为数学问题，，即对于数组prices[]，计算prices[j]-prices[i]的最大值，而且要求j>i（这一点很重要），不是单纯地计算最大值和最小值之差。
C++参考代码：
class Solution{public:int maxProfit(vector<int> &prices){vector<int>::size_type size = prices.size();if (0 == size) return 0;int min = prices[0];int result = 0;for (size_t i = 1; i < size; ++i){if (prices[i] < min) min = prices[i];else if (prices[i] – min > result) result = prices[i] – min;}return result;}};
C#参考代码：
public class Solution{(int[] prices){if (prices.Length == 0) return 0;int min = prices[0];int result = 0;for (int i = 1; i < prices.Length; ++i){if (prices[i] < min) min = prices[i];else if (prices[i] – min > result) result = prices[i] – min;}return result;}}
志在山顶的人，不会贪念山腰的风景。
