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;
    }
};

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