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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。
1 local[j] = max(local[j] + diff, global[j - 1] + max(diff, 0)); 2 global[j] = max(global[j], local[j]);
1 class Solution { 2 public: 3 int maxProfit(int k, vector<int>& prices) { 4 int result; 5 int total_times=0; 6 int profit_max = 0; 7 for(int i=1;i<prices.size();i++) 8 { 9 int diff = prices[i]-prices[i-1]; 10 if(diff>0) 11 { 12 profit_max +=diff; 13 if(((i<prices.size()-1)&&(prices[i+1]-prices[i]<=0))||(i==prices.size()-1)) 14 { 15 total_times++; 16 } 17 } 18 } 19 if(k>=total_times) return profit_max; 20 vector<int> local(k + 1, 0); 21 vector<int> global(k + 1, 0); 22 for(int i = 1; i < prices.size(); i++) 23 { 24 int diff = prices[i] - prices[i - 1]; 25 for(int j = k; j >= 1; j--) { 26 local[j] = max(local[j] + diff, global[j - 1] + max(diff, 0)); 27 global[j] = max(global[j], local[j]); 28 } 29 } 30 return global[k]; 31 } 32 };
[LeetCode]Best Time to Buy and Sell Stock IV
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原文地址：http://www.cnblogs.com/Sean-le/p/4782281.html
