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.
这道题相当简单，感觉达不到Medium的难度，只需要遍历一次数组，用一个变量记录遍历过数中的最小值，然后每次计算当前值和这个最小值之间的差值最为利润，然后每次选较大的利润来更新。当遍历完成后当前利润即为所求，代码如下：
C++ 解法：
class Solution { public: int maxProfit(vector<int>& prices) { int res = 0, buy = INT_MAX; for (int price : prices) { buy = min(buy, price); res = max(res, price - buy); } return res; } };
Java 解法：
public class Solution { public int maxProfit(int[] prices) { int res = 0, buy = Integer.MAX_VALUE; for (int price : prices) { buy = Math.min(buy, price); res = Math.max(res, price - buy); } return res; } }
本文转自博客园Grandyang的博客，原文链接：买卖股票的***时间[LeetCode] Best Time to Buy and Sell Stock ，如需转载请自行联系原博主。