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(244) 评论(0) 编辑 收藏 举报