LeetCode 121. Best Time to Buy and Sell Stock

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.

Example 1:
Input: [7, 1, 5, 3, 6, 4]
Output: 5

max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)

Example 2:
Input: [7, 6, 4, 3, 1]
Output: 0

In this case, no transaction is done, i.e. max profit = 0.

思路：
买卖股票问题，毕竟股票专业操作员，要知道卖肯定是在买之后。设置最小的索引min，最大利润max。如果当前位置的股票价格比min处小，更新min值；如果此时的利润大于max，更新max值。

代码：

class Solution {
public:int maxProfit(vector<int>& prices) {int min=0,max=0;//设置最小的索引min，最大利润maxfor(int i=1;i<prices.size();++i){if(prices[i]<prices[min]){//如果当前位置的股票价格比min处小，更新min值 min=i;}else if(prices[i]-prices[min]>max){如果此时的利润大于max，更新max值max=prices[i]-prices[min];}}return max;}
};

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