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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).
感觉此题太难，因为要求只能进行k次买卖，所以在考虑波峰，波谷的基础上要考虑买卖的次数，网上只找到了c++版的，运行了一下，可以正确运行。
class Solution { public: int maxProfit(int k, vector<int> &prices) { if(prices.empty() || k == 0) return 0; if(k >= prices.size()) return solveMaxProfit(prices); vector<int> global(k + 1, 0); vector<int> local(k + 1, 0); for(int i = 1; i < prices.size(); i++) { int diff = prices[i] - prices[i - 1]; for(int j = k; j >= 1; j--) { local[j] = max(local[j] + diff, global[j - 1] + max(diff, 0)); global[j] = max(global[j], local[j]); } } return global[k]; } private: int solveMaxProfit(vector<int> &prices) { int res = 0; for(int i = 1; i < prices.size(); i++) { int diff = prices[i] - prices[i - 1]; if(diff > 0) res += diff; } return res; } };
Best Time to Buy and Sell Stock IV
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原文地址：http://www.cnblogs.com/zhhc/p/4344821.html
