Best Time to Buy and Sell Stock III

题目要求：

Say you have an array for which the ith element is the price of a given stock on day i.
Design an algorithm to find the maximum profit. You may complete at most two transactions.
Note:You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

思路：

    因为能买2次，但第二次买不能在第一次的卖之前，找寻一个点j，将原来的price[0..n-1]分割为price[0..j]和price[j..n-1]，分别求两段的最大profit。使用两个数组，其中：
    l[i]表示 -- 截止到i下标为止，左边所做交易能够达到最大profit；
    r[i]表示 -- 截止到i下标为止，右边所做交易能够达到最大profit；
求l[i]和l[j]就演变成了Best Time to Buy and Sell Stock动态规划的过程，最后求出l[i] + r[i]的最大值即可。

代码：

#include<stdafx.h>#include <iostream>#include<vector>using namespace std;class Solution {public:int maxProfit(vector<int> &prices) {if(prices.empty())  return 0;  int n = prices.size();vector<int> l(n);vector<int> r(n);int min_t=prices[0];for (int i = 1; i < n; i++){min_t= min(min_t, prices[i]);l[i]=max(l[i-1], prices[i] - min_t);}int max_t = prices[n-1];r[n-1] = 0;for (int i = n-2; i >= 0; i--){max_t=max(max_t, prices[i]);r[i]=max(r[i+1], max_t - prices[i]);}int maxprofit = 0;for(int i = 0; i < n; i++)maxprofit=max(maxprofit, l[i] + r[i]);return maxprofit;}};void main(){int A[8]={2,3,1,7,5,4,9,5};vector<int>  prices(A,A+8);Solution s;cout<<s.maxProfit(prices)<<endl;getchar();}