C++实现LeetCode(188.买卖股票的最佳时间之四)

[LeetCode] 188.Best Time to Buy and Sell Stock IV 买卖股票的最佳时间之四

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，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;

									    }

									};