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];
        }
    }
}