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).
Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.
这道题实际上是之前那道 Best Time to Buy and Sell Stock III 买股票的最佳时间之三的一般情况的推广,还是需要用动态规划Dynamic programming来解决,具体思路如下:
这里我们需要两个递推公式来分别更新两个变量local和global,参见网友Code Ganker的博客,我们其实可以求至少k次交易的最大利润。我们定义local[i][j]为在到达第i天时最多可进行j次交易并且最后一次交易在最后一天卖出的最大利润,此为局部最优。然后我们定义global[i][j]为在到达第i天时最多可进行j次交易的最大利润,此为全局最优。它们的递推式为:
local[i][j] = max(global[i - 1][j - 1] + max(diff, 0), local[i - 1][j] + diff)
global[i][j] = max(local[i][j], global[i - 1][j]),
其中局部最优值是比较前一天并少交易一次的全局最优加上大于0的差值,和前一天的局部最优加上差值后相比,两者之中取较大值,而全局最优比较局部最优和前一天的全局最优。
但这道题还有个坑,就是如果k的值远大于prices的天数,比如k是好几百万,而prices的天数就为若干天的话,上面的DP解法就非常的没有效率,应该直接用Best Time to Buy and Sell Stock II 买股票的最佳时间之二的方法来求解,所以实际上这道题是之前的二和三的综合体,代码如下:
class Solution { public: int maxProfit(int k, vector<int> &prices) { if (prices.empty()) return 0; if (k >= prices.size()) return solveMaxProfit(prices); int g[k + 1] = {0}; int l[k + 1] = {0}; for (int i = 0; i < prices.size() - 1; ++i) { int diff = prices[i + 1] - prices[i]; for (int j = k; j >= 1; --j) { l[j] = max(g[j - 1] + max(diff, 0), l[j] + diff); g[j] = max(g[j], l[j]); } } return g[k]; } int solveMaxProfit(vector<int> &prices) { int res = 0; for (int i = 1; i < prices.size(); ++i) { if (prices[i] - prices[i - 1] > 0) { res += prices[i] - prices[i - 1]; } } return res; } };
本文转自博客园Grandyang的博客,原文链接:买卖股票的最佳时间之四[LeetCode] Best Time to Buy and Sell Stock IV ,如需转载请自行联系原博主。