[LeetCode]Best Time to Buy and Sell Stock IV

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.

这题难度明显增大。是Best Time to Buy and Sell Stock II和Best Time to Buy and Sell Stock III两道题的结合。

参考博文，主要是两个递推公式。

local[j] = max(local[j] + diff, global[j - 1] + max(diff, 0));
global[j] = max(global[j], local[j]);

class Solution {
public:
    int maxProfit(int k, vector<int>& prices) {
        int result;
        int total_times=0;
        int profit_max = 0;
        for(int i=1;i<prices.size();i++)
        {
            int diff = prices[i]-prices[i-1];
            if(diff>0)
            {
                profit_max +=diff;
                if(((i<prices.size()-1)&&(prices[i+1]-prices[i]<=0))||(i==prices.size()-1))
                {
                    total_times++;
                }
            }
        }
        if(k>=total_times) return profit_max;
        vector<int> local(k + 1, 0);
        vector<int> global(k + 1, 0);
        for(int i = 1; i < prices.size(); i++) 
        {
            int diff = prices[i] - prices[i - 1];
            for(int j = k; j >= 1; j--) {
                local[j] = max(local[j] + diff, global[j - 1] + max(diff, 0));
                global[j] = max(global[j], local[j]);
            }
        }
        return global[k];
    }
};