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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).
和Best Time to Buy and Sell Stock很接近，需要维护两个vector使得两次交易一次在区间【0-i】，一次在区间【i，n-1】。
每个区间内的求解最大收益的方法和Best Time to Buy and Sell Stock几乎一样。
取两个区间收益和最大的的结果为最终结果。
1 class Solution { 2 public: 3 int maxProfit(vector<int>& prices) { 4 if (prices.size()<2) return 0; 5 int result=0; 6 int n = prices.size(); 7 vector<int> left(n,0); 8 vector<int> right(n,0); 9 int cur_min = prices[0]; 10 int cur_result = 0; 11 for(int i=1;i<n;i++) 12 { 13 cur_result = max(cur_result,prices[i]-cur_min); 14 left[i] = cur_result; 15 cur_min = min(cur_min,prices[i]); 16 } 17 int cur_max = prices[n-1]; 18 cur_result = 0; 19 for(int i=n-2;i>0;i--) 20 { 21 cur_result = max(cur_result,cur_max-prices[i]); 22 right[i] = cur_result; 23 cur_max = max(cur_max,prices[i]); 24 } 25 for(int i=1;i<n;i++) 26 { 27 result = max(result,left[i]+right[i]); 28 } 29 return result; 30 } 31 };
[LeetCode]Best Time to Buy and Sell Stock III
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原文地址：http://www.cnblogs.com/Sean-le/p/4782008.html
