121. Best Time to Buy and Sell Stock

1. 问题描述
   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: 5max. 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: 0In this case, no transaction is done, i.e. max profit = 0.

2.解决思路
动态规划的思路。设buy[i]为prices[0…i]的买入价最小的值，sell[i]为prices[0…i]的卖出价的最大值。
则状态转移方程为:
buy[i] = max(buy[i-1],-prices[i])
sell[i] = max(sell[i-1],prices[i]+buy[i])
输出结果为sell[n-1]
但是这里显然可以做空间压缩，只要常数个变量就能实现动态规划的算法。

3.代码

class Solution {public:    int maxProfit(vector<int>& prices) {        if (prices.size() == 0)            return 0;        if (prices.size() == 0)            return 0;        int buy1 = INT_MIN ,sell1 = INT_MIN , buy2 = INT_MIN, sell2 = INT_MIN;        for (int i = 0 ; i < prices.size(); ++i) {            buy1 = max(buy1,-prices[i]);            sell1 = max(sell1,prices[i]+buy1);        }        return sell1;    }};