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.

 public class Solution {
     public int maxProfit(int k, int[] prices) {
         if(prices == null || prices.length == 0) return 0;
         int len = prices.length;
         if (k >= len / 2) return quickSolve(prices);
         int[][] dp = new int[k + 1][len];
         for (int i = 1; i <= k; i++) {
             int tmpMax = -prices[0];
             for(int j = 1; j < len; j ++){
                 dp[i][j] = Math.max(dp[i][j - 1], prices[j] + tmpMax);
                 tmpMax = Math.max(tmpMax, dp[i - 1][j - 1] - prices[j]);
             }
         }
         return dp[k][len - 1];
     }

     public int quickSolve(int[] prices){
         int result = 0;
         for(int i = 1; i < prices.length; i ++){
             if(prices[i] - prices[i - 1] > 0) result += prices[i] - prices[i - 1];
         }
         return result;
     }
 }

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.

 public class Solution {
     public int maxProfit(int k, int[] prices) {
         if(prices == null || prices.length < 2 || k == 0) return 0;
         int len = prices.length;
         if(k * 2 >= len){//actually we can do as many transactions as we want
             int result = 0;
             for(int i = 1; i < len; i ++){
                 if(prices[i] - prices[i - 1] > 0) result += prices[i] - prices[i - 1];
             }
             return result;
         }else{//transactions time is limited
             int[][] dp = new int[k + 1][len];
             for(int i = 1; i <= k; i ++){
                 int MaxPre = -prices[0];
                 for(int j = 1; j < len; j ++){
                     dp[i][j] = Math.max(dp[i][j - 1], MaxPre + prices[j]);
                     MaxPre = Math.max(MaxPre, dp[i - 1][j - 1] - prices[j]);
                 }
             }
             return dp[k][len - 1];
         }
     }
 }