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.

 

 1 public class Solution {
 2     public int maxProfit(int k, int[] prices) {
 3         if(prices == null || prices.length == 0) return 0;
 4         int len = prices.length;
 5         if (k >= len / 2) return quickSolve(prices);
 6         int[][] dp = new int[k + 1][len];
 7         for (int i = 1; i <= k; i++) {
 8             int tmpMax = -prices[0];
 9             for(int j = 1; j < len; j ++){
10                 dp[i][j] = Math.max(dp[i][j - 1], prices[j] + tmpMax);
11                 tmpMax = Math.max(tmpMax, dp[i - 1][j - 1] - prices[j]);
12             }
13         }
14         return dp[k][len - 1];
15     }
16     
17     public int quickSolve(int[] prices){
18         int result = 0;
19         for(int i = 1; i < prices.length; i ++){
20             if(prices[i] - prices[i - 1] > 0) result += prices[i] - prices[i - 1];
21         }
22         return result;
23     }
24 }

tmpMax means the maximum profit of just doing at most i-1 transactions, using at most first j-1 prices, and buying the stock at price[j] - this is used for the next loop.

 

 1 public class Solution {
 2     public int maxProfit(int k, int[] prices) {
 3         if(prices == null || prices.length < 2 || k == 0) return 0;
 4         int len = prices.length;
 5         if(k * 2 >= len){//actually we can do as many transactions as we want
 6             int result = 0;
 7             for(int i = 1; i < len; i ++){
 8                 if(prices[i] - prices[i - 1] > 0) result += prices[i] - prices[i - 1];
 9             }
10             return result;
11         }else{//transactions time is limited
12             int[][] dp = new int[k + 1][len];
13             for(int i = 1; i <= k; i ++){
14                 int MaxPre = -prices[0];
15                 for(int j = 1; j < len; j ++){
16                     dp[i][j] = Math.max(dp[i][j - 1], MaxPre + prices[j]);
17                     MaxPre = Math.max(MaxPre, dp[i - 1][j - 1] - prices[j]);
18                 }
19             }
20             return dp[k][len - 1];
21         }
22     }
23 }

 

posted on 2015-03-25 14:42  Step-BY-Step  阅读(243)  评论(0)  编辑  收藏  举报

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