LeetCode: 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.

给出一组数，其中下标为i的数表示第i天股票的价格，可以买入和卖出股票（要求必须先买入再卖出且只能交易一次），求买卖过程中的最大收益。
如果是枚举的话，算法复杂度o(n^2)，很容易超时。
用动态规划法，能将算法的时间复杂度减小到o(n)。将最大收益初始化为0，从前到后遍历数组，记录当前出现过的最低价格，作为买入价格。同时，计算以当天价格出售时的收益，如果比最大收益大，则替换最大收益，否则最大收益不变。状态转移方程：
maxProfit=max{maxProfit，prices[i]-curMinprice}
Accepted的代码：

class Solution {public:    int maxProfit(vector<int>& prices) {        if(prices.size()==0||prices.size()==1) return 0;        int max_profit=0;//最大收益        int cur_min_price=prices[0];//当前出现过的最低价格        for(int i=1;i<prices.size();i++)        {            if(prices[i]<cur_min_price) cur_min_price=prices[i];            if(prices[i]-cur_min_price>max_profit) max_profit=prices[i]-cur_min_price;        }        return max_profit;    }};