标签：
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.
代码：
class Solution { public: int maxProfit(vector<int>& prices) { if (prices.size()==0) return 0; int min_price = prices[0], max_profit = 0; for ( size_t i = 0; i < prices.size(); ++i ) { min_price = std::min(min_price, prices[i]); max_profit = std::max(max_profit, prices[i]-min_price); } return max_profit; } };
tips：
Greedy
核心在于维护两个变量：
min_price: 算上当前元素的最小值
max_profit：算上当前元素的最大利润
走完所有元素，可以获得max_profit
【Best Time to Buy and Sell Stock】cpp
标签：
原文地址：http://www.cnblogs.com/xbf9xbf/p/4539811.html
