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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 k 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：用状态存储至当前日为止第jth buy/sell的最大利润。到了第二天，我们可以按j从大到小（因为j大的新状态依赖于之前j小的状态），修改这个状态。
class Solution { public: int maxProfit(int k, vector<int>& prices) { int dates = prices.size(); if(dates <= 1 || k == 0) return 0; if (k >= prices.size()) return maxProfit2(prices); //unlimited transaction vector<int> release(k,0); //sell stock vector<int> hold(k,INT_MIN); //buy stock for(int i = 0; i < dates; i++){ for(int j = k-1; j > 0; j--){ release[j] = max(release[j], hold[j]+prices[i]); //jth sell happen at ith day hold[j]=max(hold[j], release[j-1]-prices[i]); //jth buy happen at ith day } release[0] = max(release[0], hold[0]+prices[i]); hold[0] = max(hold[0],-prices[i]); } return release[k-1]; } int maxProfit2(vector<int> &prices) { int profit = 0; for (int i=0; i<(int)prices.size()-1; i++) { if (prices[i+1] > prices[i]) profit += prices[i+1] - prices[i]; } return profit; } };
188. Best Time to Buy and Sell Stock IV (Array; DP)
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原文地址：http://www.cnblogs.com/qionglouyuyu/p/5050075.html
