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. （Hard）
Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
分析：
最多购买两次股票且不能同时拥有，所以买卖两次肯定分布在前后两个不同的区间。
设dp(i) = 区间[0,1,2...i]的最大利润 + 区间[i,i+1,....n-1]的最大利润,那么本题的最大利润 = max{dp[0],dp[1],dp[2],...,dp[n-1]}.
而对于式子中两个区间内分别只能有一次买卖，这是Best Time to Buy and Sell Stock I的问题。
代码：
1 class Solution { 2 public: 3 int maxProfit(vector<int> &prices) { 4 int len = prices.size(); 5 if(len <= 1) { 6 return 0; 7 } 8 int maxFromHead[len]; 9 maxFromHead[0] = 0; 10 int minprice = prices[0], maxprofit = 0; 11 for(int i = 1; i < len; i++) { 12 minprice = min(prices[i-1], minprice); 13 if(maxprofit < prices[i] - minprice) 14 maxprofit = prices[i] - minprice; 15 maxFromHead[i] = maxprofit; 16 } 17 int maxprice = prices[len - 1]; 18 int res = maxFromHead[len-1]; 19 maxprofit = 0; 20 for(int i = len-2; i >=0; i--) { 21 maxprice = max(maxprice, prices[i+1]); 22 if(maxprofit < maxprice - prices[i]) 23 maxprofit = maxprice - prices[i]; 24 if(res < maxFromHead[i] + maxprofit) 25 res = maxFromHead[i] + maxprofit; 26 } 27 return res; 28 } 29 };