188. Best Time to Buy and Sell Stock IV

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.

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题意：

与123. Best Time to Buy and Sell Stock III不同，可以买卖k次。

思路：

实质上就是122. Best Time to Buy and Sell Stock II和123. Best Time to Buy and Sell Stock III的结合。

大致上与前一题相同，都是找状态转移方程。只不过需要注意的是，如果k>n/2，这就是122. Best Time to Buy and Sell Stock II的情况，那么就可以自由交易，即可以买卖相间，所以对于有利益的，我们就进行一次买卖，亏损的就不买卖了，利润为0。否则对于一般的情况，可以用123. Best Time to Buy and Sell Stock III的思路。

此外特别注意的是：k可能很大，所以最好不要用profit数组，用一个vector来保存，才不会runtime error。

代码：

class Solution {
public:
    int maxProfit(int k, vector<int>& prices) {
        if(k<=0)
            return 0;
        //int profit[2*k];
        vector<int>profit;
        int n=prices.size();
        int maxprofit=0;
        if(k>n/2)
        {
            for(int i=1;i<n;i++)
                maxprofit+=max(prices[i]-prices[i-1],0);
            return maxprofit;
        }
        for(int i=0;i<2*k;i++)
        {
            profit.push_back(INT_MIN);
            profit.push_back(0);
        }
        for(int i=0;i<n;i++)
        {
            for(int j=2*k-1;j>=0;j--)
            {
                if(j!=0)
                {
                    if(j%2==1)
                        profit[j]=max(profit[j],profit[j-1]+prices[i]);
                    else
                        profit[j]=max(profit[j],profit[j-1]-prices[i]);
                }
                else
                    profit[j]=max(profit[j],-prices[i]);
            }
        }
        for(int i=1;i<=2*k-1;i+=2)
        {
            maxprofit=max(profit[i],maxprofit);
        }
        return maxprofit;
    }
};