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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).
思路：把要求的东西profits设成状态。 从左往右，从右往左两次scan赋值状态，然后再扫描一次状态，所以时间复杂度O(n) *3
class Solution { public: int maxProfit(vector<int> &prices) { if(prices.empty()) return 0; int size = prices.size(); int maxProfit = 0; int* profits = new int [size]; //表示从0到i的最大利润 int* profits_from_last = new int [size]; //表示从i到size-1的最大利润 //initial status profits[0] = 0; int minValue = prices[0]; //scan from start for(int i = 1; i< size; i++) { if(prices[i] < minValue) { minValue = prices[i]; profits[i] = profits[i-1]; } else { profits[i] = max(prices[i]-minValue,profits[i-1]); } } //initial status profits_from_last[size-1] = 0; int maxValue = prices[size-1]; //scan from last for(int i = size-2; i >=0 ; i--) { if(prices[i] > maxValue) { maxValue = prices[i]; profits_from_last[i] = profits_from_last[i+1]; } else { profits_from_last[i] = max(maxValue-prices[i],profits_from_last[i+1]); } int profit = profits[i] + profits_from_last[i]; if(profit>maxProfit) { maxProfit = profit; } } return maxProfit; } };
123. Best Time to Buy and Sell Stock III (Array; DP)
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原文地址：http://www.cnblogs.com/qionglouyuyu/p/4878356.html
