Say you have an array for which the i^th 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.

class Solution {
public:
    int maxProfit(int k, vector<int> &prices) {
        if (k >= prices.size())
            return maxProfit2(prices);
            
        vector<int> release(k+1);
        vector<int> hold(k, INT_MIN);
        
        for (auto p: prices) {
            for (int i=0; i<k; i++) {
                release[i] = max(release[i], hold[i]+p);
                hold[i] = max(hold[i], release[i+1]-p);
            }
        }
        
        return release[0];
    }
    
    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;
    }
};
 

此題思路借鑑之以下leetcode討論：

該討論是針對交易次數為2次的的情況。

但思路是一樣的。我將其擴充到任意次數的情況。

需要注意的事，test case中，有一個非常大的k值，直接會讓記憶體分配失敗。

如何處理該種情況呢。 當k值超過prices值的個數時，此時，可以把問題轉換為交易數次不限的情況。即

Best Time to Buy and Sell Stock II

另一篇博文也很好的介紹了這個問題的解法。這兩者的思路其實是一樣的。